Variance of $XY(1-Y)$ in terms of the means and variances $X$ and $Y$












2












$begingroup$


Consider two independent random variables X and Y. X has some distribution with mean $mu_X$ and variance $sigma^2_X$. Y has some distribution with mean $mu_Y$ and variance $sigma^2_Y$. I want to take the variance of the quantity $XY(1-Y)$. Is the correct answer:
$$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]?$$
I just did some kind of weird product rule but with variances and I'm not sure that's at all legit.



The reason I want to know the variance of this quantity is that it is a measure of the real noise in an experimental system I am working with. I am an experimental physicist and my math skills are weak at best. You might recognize the form $XY(1-Y)$ from the binomial distribution: $Np(1-p)$. My experiment does not follow a binomial distribution exactly, it follows a Poisson binomial distribution but with varying $N$ and $p$ for each shot of the experiment. So I need to know the variance of the quantity $XY(1-Y)$ when I know the variances and expectation values of $X$ and $Y$.



I'm trying to work this particular detail out to answer this question:
Repeated binomial processes with different probabilities and number of trials .






probability-theory variance







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – quid♦
    Dec 24 '18 at 14:55






  • 1




    $begingroup$
    The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
    $endgroup$
    – Did
    Dec 25 '18 at 18:08
















2












$begingroup$


Consider two independent random variables X and Y. X has some distribution with mean $mu_X$ and variance $sigma^2_X$. Y has some distribution with mean $mu_Y$ and variance $sigma^2_Y$. I want to take the variance of the quantity $XY(1-Y)$. Is the correct answer:
$$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]?$$
I just did some kind of weird product rule but with variances and I'm not sure that's at all legit.



The reason I want to know the variance of this quantity is that it is a measure of the real noise in an experimental system I am working with. I am an experimental physicist and my math skills are weak at best. You might recognize the form $XY(1-Y)$ from the binomial distribution: $Np(1-p)$. My experiment does not follow a binomial distribution exactly, it follows a Poisson binomial distribution but with varying $N$ and $p$ for each shot of the experiment. So I need to know the variance of the quantity $XY(1-Y)$ when I know the variances and expectation values of $X$ and $Y$.



I'm trying to work this particular detail out to answer this question:
Repeated binomial processes with different probabilities and number of trials .






probability-theory variance







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – quid♦
    Dec 24 '18 at 14:55






  • 1




    $begingroup$
    The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
    $endgroup$
    – Did
    Dec 25 '18 at 18:08














2












2








2





$begingroup$


Consider two independent random variables X and Y. X has some distribution with mean $mu_X$ and variance $sigma^2_X$. Y has some distribution with mean $mu_Y$ and variance $sigma^2_Y$. I want to take the variance of the quantity $XY(1-Y)$. Is the correct answer:
$$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]?$$
I just did some kind of weird product rule but with variances and I'm not sure that's at all legit.



The reason I want to know the variance of this quantity is that it is a measure of the real noise in an experimental system I am working with. I am an experimental physicist and my math skills are weak at best. You might recognize the form $XY(1-Y)$ from the binomial distribution: $Np(1-p)$. My experiment does not follow a binomial distribution exactly, it follows a Poisson binomial distribution but with varying $N$ and $p$ for each shot of the experiment. So I need to know the variance of the quantity $XY(1-Y)$ when I know the variances and expectation values of $X$ and $Y$.



I'm trying to work this particular detail out to answer this question:
Repeated binomial processes with different probabilities and number of trials .






probability-theory variance







share|cite|improve this question











$endgroup$




Consider two independent random variables X and Y. X has some distribution with mean $mu_X$ and variance $sigma^2_X$. Y has some distribution with mean $mu_Y$ and variance $sigma^2_Y$. I want to take the variance of the quantity $XY(1-Y)$. Is the correct answer:
$$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]?$$
I just did some kind of weird product rule but with variances and I'm not sure that's at all legit.



The reason I want to know the variance of this quantity is that it is a measure of the real noise in an experimental system I am working with. I am an experimental physicist and my math skills are weak at best. You might recognize the form $XY(1-Y)$ from the binomial distribution: $Np(1-p)$. My experiment does not follow a binomial distribution exactly, it follows a Poisson binomial distribution but with varying $N$ and $p$ for each shot of the experiment. So I need to know the variance of the quantity $XY(1-Y)$ when I know the variances and expectation values of $X$ and $Y$.



I'm trying to work this particular detail out to answer this question:
Repeated binomial processes with different probabilities and number of trials .






probability-theory variance




probability-theory variance






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 18:09









Did

249k23228467




249k23228467










asked Dec 23 '18 at 5:42









SabrinaChoiceSabrinaChoice

247




247








  • 1




    $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – quid♦
    Dec 24 '18 at 14:55






  • 1




    $begingroup$
    The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
    $endgroup$
    – Did
    Dec 25 '18 at 18:08














  • 1




    $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – quid♦
    Dec 24 '18 at 14:55






  • 1




    $begingroup$
    The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
    $endgroup$
    – Did
    Dec 25 '18 at 18:08








1




1




$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– quid♦
Dec 24 '18 at 14:55




$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– quid♦
Dec 24 '18 at 14:55




1




1




$begingroup$
The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
$endgroup$
– Did
Dec 25 '18 at 18:08




$begingroup$
The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
$endgroup$
– Did
Dec 25 '18 at 18:08










1 Answer
1






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2












$begingroup$

$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    $endgroup$
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    $begingroup$
    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    $endgroup$
    – user21820
    Dec 24 '18 at 15:02






  • 1




    $begingroup$
    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    $endgroup$
    – drhab
    Dec 25 '18 at 16:44








  • 1




    $begingroup$
    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    $endgroup$
    – drhab
    Dec 26 '18 at 8:15












  • $begingroup$
    @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    $endgroup$
    – SabrinaChoice
    Jan 4 at 6:52














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1 Answer
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1 Answer
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active

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$begingroup$

$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    $endgroup$
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    $begingroup$
    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    $endgroup$
    – user21820
    Dec 24 '18 at 15:02






  • 1




    $begingroup$
    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    $endgroup$
    – drhab
    Dec 25 '18 at 16:44








  • 1




    $begingroup$
    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    $endgroup$
    – drhab
    Dec 26 '18 at 8:15












  • $begingroup$
    @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    $endgroup$
    – SabrinaChoice
    Jan 4 at 6:52


















2












$begingroup$

$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    $endgroup$
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    $begingroup$
    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    $endgroup$
    – user21820
    Dec 24 '18 at 15:02






  • 1




    $begingroup$
    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    $endgroup$
    – drhab
    Dec 25 '18 at 16:44








  • 1




    $begingroup$
    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    $endgroup$
    – drhab
    Dec 26 '18 at 8:15












  • $begingroup$
    @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    $endgroup$
    – SabrinaChoice
    Jan 4 at 6:52
















2












2








2





$begingroup$

$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).






share|cite|improve this answer











$endgroup$



$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 8:06

























answered Dec 24 '18 at 13:11









drhabdrhab

104k545136




104k545136












  • $begingroup$
    This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    $endgroup$
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    $begingroup$
    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    $endgroup$
    – user21820
    Dec 24 '18 at 15:02






  • 1




    $begingroup$
    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    $endgroup$
    – drhab
    Dec 25 '18 at 16:44








  • 1




    $begingroup$
    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    $endgroup$
    – drhab
    Dec 26 '18 at 8:15












  • $begingroup$
    @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    $endgroup$
    – SabrinaChoice
    Jan 4 at 6:52




















  • $begingroup$
    This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    $endgroup$
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    $begingroup$
    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    $endgroup$
    – user21820
    Dec 24 '18 at 15:02






  • 1




    $begingroup$
    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    $endgroup$
    – drhab
    Dec 25 '18 at 16:44








  • 1




    $begingroup$
    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    $endgroup$
    – drhab
    Dec 26 '18 at 8:15












  • $begingroup$
    @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    $endgroup$
    – SabrinaChoice
    Jan 4 at 6:52


















$begingroup$
This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
$endgroup$
– SabrinaChoice
Dec 24 '18 at 14:47






$begingroup$
This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
$endgroup$
– SabrinaChoice
Dec 24 '18 at 14:47






1




1




$begingroup$
@SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
$endgroup$
– user21820
Dec 24 '18 at 15:02




$begingroup$
@SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
$endgroup$
– user21820
Dec 24 '18 at 15:02




1




1




$begingroup$
@SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
$endgroup$
– drhab
Dec 25 '18 at 16:44






$begingroup$
@SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
$endgroup$
– drhab
Dec 25 '18 at 16:44






1




1




$begingroup$
In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
$endgroup$
– drhab
Dec 26 '18 at 8:15






$begingroup$
In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
$endgroup$
– drhab
Dec 26 '18 at 8:15














$begingroup$
@drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
$endgroup$
– SabrinaChoice
Jan 4 at 6:52






$begingroup$
@drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
$endgroup$
– SabrinaChoice
Jan 4 at 6:52




















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