Computing marginal PDF given joint pdf












0












$begingroup$



Let the joint PDF of $(X, Y)$ be given by



$$f(x, y) = frac{2}{x} e^{-2x}, hspace{1cm} x>0, 0 leq y< x. $$



a) Find the marginal PDF of $X$ and find $mathbb{E}[X]$



b) Find the conditional pdf of $Y$ given $X = x$



c) Compute $text{Cov}(X, Y)$.




MY try:



a)



$$f_{X}(x) = int_{-infty}^{infty} f(x, y) mathop{dy} = int_{0}^{x} frac{2}{x}e^{-2x} mathop{dy}$$



$$ = boxed{2e^{-2x} hspace{1cm} text{ for } x > 0}$$



Also,
$$mathbb{E}[X] = int_{-infty}^{infty} x cdot f_{X}(x) mathop{dx} = int_{0}^{infty} 2xe^{-2x} mathop{dx} = boxed{frac{1}{2}}$$



b)



$$f_{Y mid X}(y mid x) = frac{f(x, y)}{f_{X}(x)} = boxed{frac{1}{x} text{ for } x > 0, x neq 0}$$



c)



$$text{Cov}(X,Y) = mathbb{E}[XY] - mathbb{E}[X]cdot mathbb{E}[Y]. $$



First, compute $mathbb{E}[Y]$ as follows:



$$mathbb{E}[Y] = mathbb{E}[mathbb{E}[Y mid X]] = mathbb{E}[frac{X}{2}] = frac{1}{2} cdot mathbb{E}[X] = frac{1}{4}$$



Now compute $mathbb{E}[XY]$:



$$mathbb{E}[XY] = int_{0}^{infty} int_{0}^{x} 2ye^{-2x} mathop{dy} mathop{dx} $$



$$ = int_{0}^{infty} x^{2}e^{-2x} mathop{dx}$$



$$= mathbb{E}[X^{2}]/2.$$



Recall we have $text{Var}(X) = 1/lambda^{2}$, which means $mathbb{E}[X^{2}] = mathbb{E}[X]^{2} + text{Var}(X) = frac{2}{lambda^{2}}$. Therefore,



$$text{Cov}(X, Y) = frac{1}{4} - frac{1}{8} = boxed{frac{1}{8}} $$




Are all of these answers correct? I'm unsure about the domains to be honest for (a) and (b).






probability probability-theory probability-distributions







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$endgroup$












  • $begingroup$
    (a) $E(X)$ is wrong. It should be $int_0^infty 2xe^{-2x}dx=frac{1}{2}$.
    $endgroup$
    – herb steinberg
    Dec 13 '18 at 3:12
















0












$begingroup$



Let the joint PDF of $(X, Y)$ be given by



$$f(x, y) = frac{2}{x} e^{-2x}, hspace{1cm} x>0, 0 leq y< x. $$



a) Find the marginal PDF of $X$ and find $mathbb{E}[X]$



b) Find the conditional pdf of $Y$ given $X = x$



c) Compute $text{Cov}(X, Y)$.




MY try:



a)



$$f_{X}(x) = int_{-infty}^{infty} f(x, y) mathop{dy} = int_{0}^{x} frac{2}{x}e^{-2x} mathop{dy}$$



$$ = boxed{2e^{-2x} hspace{1cm} text{ for } x > 0}$$



Also,
$$mathbb{E}[X] = int_{-infty}^{infty} x cdot f_{X}(x) mathop{dx} = int_{0}^{infty} 2xe^{-2x} mathop{dx} = boxed{frac{1}{2}}$$



b)



$$f_{Y mid X}(y mid x) = frac{f(x, y)}{f_{X}(x)} = boxed{frac{1}{x} text{ for } x > 0, x neq 0}$$



c)



$$text{Cov}(X,Y) = mathbb{E}[XY] - mathbb{E}[X]cdot mathbb{E}[Y]. $$



First, compute $mathbb{E}[Y]$ as follows:



$$mathbb{E}[Y] = mathbb{E}[mathbb{E}[Y mid X]] = mathbb{E}[frac{X}{2}] = frac{1}{2} cdot mathbb{E}[X] = frac{1}{4}$$



Now compute $mathbb{E}[XY]$:



$$mathbb{E}[XY] = int_{0}^{infty} int_{0}^{x} 2ye^{-2x} mathop{dy} mathop{dx} $$



$$ = int_{0}^{infty} x^{2}e^{-2x} mathop{dx}$$



$$= mathbb{E}[X^{2}]/2.$$



Recall we have $text{Var}(X) = 1/lambda^{2}$, which means $mathbb{E}[X^{2}] = mathbb{E}[X]^{2} + text{Var}(X) = frac{2}{lambda^{2}}$. Therefore,



$$text{Cov}(X, Y) = frac{1}{4} - frac{1}{8} = boxed{frac{1}{8}} $$




Are all of these answers correct? I'm unsure about the domains to be honest for (a) and (b).






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    (a) $E(X)$ is wrong. It should be $int_0^infty 2xe^{-2x}dx=frac{1}{2}$.
    $endgroup$
    – herb steinberg
    Dec 13 '18 at 3:12














0












0








0





$begingroup$



Let the joint PDF of $(X, Y)$ be given by



$$f(x, y) = frac{2}{x} e^{-2x}, hspace{1cm} x>0, 0 leq y< x. $$



a) Find the marginal PDF of $X$ and find $mathbb{E}[X]$



b) Find the conditional pdf of $Y$ given $X = x$



c) Compute $text{Cov}(X, Y)$.




MY try:



a)



$$f_{X}(x) = int_{-infty}^{infty} f(x, y) mathop{dy} = int_{0}^{x} frac{2}{x}e^{-2x} mathop{dy}$$



$$ = boxed{2e^{-2x} hspace{1cm} text{ for } x > 0}$$



Also,
$$mathbb{E}[X] = int_{-infty}^{infty} x cdot f_{X}(x) mathop{dx} = int_{0}^{infty} 2xe^{-2x} mathop{dx} = boxed{frac{1}{2}}$$



b)



$$f_{Y mid X}(y mid x) = frac{f(x, y)}{f_{X}(x)} = boxed{frac{1}{x} text{ for } x > 0, x neq 0}$$



c)



$$text{Cov}(X,Y) = mathbb{E}[XY] - mathbb{E}[X]cdot mathbb{E}[Y]. $$



First, compute $mathbb{E}[Y]$ as follows:



$$mathbb{E}[Y] = mathbb{E}[mathbb{E}[Y mid X]] = mathbb{E}[frac{X}{2}] = frac{1}{2} cdot mathbb{E}[X] = frac{1}{4}$$



Now compute $mathbb{E}[XY]$:



$$mathbb{E}[XY] = int_{0}^{infty} int_{0}^{x} 2ye^{-2x} mathop{dy} mathop{dx} $$



$$ = int_{0}^{infty} x^{2}e^{-2x} mathop{dx}$$



$$= mathbb{E}[X^{2}]/2.$$



Recall we have $text{Var}(X) = 1/lambda^{2}$, which means $mathbb{E}[X^{2}] = mathbb{E}[X]^{2} + text{Var}(X) = frac{2}{lambda^{2}}$. Therefore,



$$text{Cov}(X, Y) = frac{1}{4} - frac{1}{8} = boxed{frac{1}{8}} $$




Are all of these answers correct? I'm unsure about the domains to be honest for (a) and (b).






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$





Let the joint PDF of $(X, Y)$ be given by



$$f(x, y) = frac{2}{x} e^{-2x}, hspace{1cm} x>0, 0 leq y< x. $$



a) Find the marginal PDF of $X$ and find $mathbb{E}[X]$



b) Find the conditional pdf of $Y$ given $X = x$



c) Compute $text{Cov}(X, Y)$.




MY try:



a)



$$f_{X}(x) = int_{-infty}^{infty} f(x, y) mathop{dy} = int_{0}^{x} frac{2}{x}e^{-2x} mathop{dy}$$



$$ = boxed{2e^{-2x} hspace{1cm} text{ for } x > 0}$$



Also,
$$mathbb{E}[X] = int_{-infty}^{infty} x cdot f_{X}(x) mathop{dx} = int_{0}^{infty} 2xe^{-2x} mathop{dx} = boxed{frac{1}{2}}$$



b)



$$f_{Y mid X}(y mid x) = frac{f(x, y)}{f_{X}(x)} = boxed{frac{1}{x} text{ for } x > 0, x neq 0}$$



c)



$$text{Cov}(X,Y) = mathbb{E}[XY] - mathbb{E}[X]cdot mathbb{E}[Y]. $$



First, compute $mathbb{E}[Y]$ as follows:



$$mathbb{E}[Y] = mathbb{E}[mathbb{E}[Y mid X]] = mathbb{E}[frac{X}{2}] = frac{1}{2} cdot mathbb{E}[X] = frac{1}{4}$$



Now compute $mathbb{E}[XY]$:



$$mathbb{E}[XY] = int_{0}^{infty} int_{0}^{x} 2ye^{-2x} mathop{dy} mathop{dx} $$



$$ = int_{0}^{infty} x^{2}e^{-2x} mathop{dx}$$



$$= mathbb{E}[X^{2}]/2.$$



Recall we have $text{Var}(X) = 1/lambda^{2}$, which means $mathbb{E}[X^{2}] = mathbb{E}[X]^{2} + text{Var}(X) = frac{2}{lambda^{2}}$. Therefore,



$$text{Cov}(X, Y) = frac{1}{4} - frac{1}{8} = boxed{frac{1}{8}} $$




Are all of these answers correct? I'm unsure about the domains to be honest for (a) and (b).






probability probability-theory probability-distributions




probability probability-theory probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 3:19







joseph

















asked Dec 13 '18 at 2:53









josephjoseph

496111




496111












  • $begingroup$
    (a) $E(X)$ is wrong. It should be $int_0^infty 2xe^{-2x}dx=frac{1}{2}$.
    $endgroup$
    – herb steinberg
    Dec 13 '18 at 3:12


















  • $begingroup$
    (a) $E(X)$ is wrong. It should be $int_0^infty 2xe^{-2x}dx=frac{1}{2}$.
    $endgroup$
    – herb steinberg
    Dec 13 '18 at 3:12
















$begingroup$
(a) $E(X)$ is wrong. It should be $int_0^infty 2xe^{-2x}dx=frac{1}{2}$.
$endgroup$
– herb steinberg
Dec 13 '18 at 3:12




$begingroup$
(a) $E(X)$ is wrong. It should be $int_0^infty 2xe^{-2x}dx=frac{1}{2}$.
$endgroup$
– herb steinberg
Dec 13 '18 at 3:12










1 Answer
1






active

oldest

votes


















1












$begingroup$

Guide:




  • $E[X], E[Y], E[XY]$ should all be just numbers.


  • $X$ follows an exponential distribution, the expectation is $frac12$.


  • begin{align}
    E[XY] &= int_0^infty int_0^x 2y e^{-2x} , dy , dx \
    &=frac12int_0^infty 2x^2 e^{-2x}, dx \
    &= frac12 E[X^2]
    end{align}







share|cite|improve this answer









$endgroup$













  • $begingroup$
    I see. If i wanted to compute the integral $int x cdot f_{X}(x)$, what would the bounds be ?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:11










  • $begingroup$
    from $0$ to $infty$, that's where $f_X(x)$ is positive.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:12












  • $begingroup$
    ok let me try again
    $endgroup$
    – joseph
    Dec 13 '18 at 3:12










  • $begingroup$
    ok can you check again?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:20










  • $begingroup$
    seems fine to me.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:23











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

Guide:




  • $E[X], E[Y], E[XY]$ should all be just numbers.


  • $X$ follows an exponential distribution, the expectation is $frac12$.


  • begin{align}
    E[XY] &= int_0^infty int_0^x 2y e^{-2x} , dy , dx \
    &=frac12int_0^infty 2x^2 e^{-2x}, dx \
    &= frac12 E[X^2]
    end{align}







share|cite|improve this answer









$endgroup$













  • $begingroup$
    I see. If i wanted to compute the integral $int x cdot f_{X}(x)$, what would the bounds be ?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:11










  • $begingroup$
    from $0$ to $infty$, that's where $f_X(x)$ is positive.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:12












  • $begingroup$
    ok let me try again
    $endgroup$
    – joseph
    Dec 13 '18 at 3:12










  • $begingroup$
    ok can you check again?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:20










  • $begingroup$
    seems fine to me.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:23
















1












$begingroup$

Guide:




  • $E[X], E[Y], E[XY]$ should all be just numbers.


  • $X$ follows an exponential distribution, the expectation is $frac12$.


  • begin{align}
    E[XY] &= int_0^infty int_0^x 2y e^{-2x} , dy , dx \
    &=frac12int_0^infty 2x^2 e^{-2x}, dx \
    &= frac12 E[X^2]
    end{align}







share|cite|improve this answer









$endgroup$













  • $begingroup$
    I see. If i wanted to compute the integral $int x cdot f_{X}(x)$, what would the bounds be ?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:11










  • $begingroup$
    from $0$ to $infty$, that's where $f_X(x)$ is positive.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:12












  • $begingroup$
    ok let me try again
    $endgroup$
    – joseph
    Dec 13 '18 at 3:12










  • $begingroup$
    ok can you check again?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:20










  • $begingroup$
    seems fine to me.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:23














1












1








1





$begingroup$

Guide:




  • $E[X], E[Y], E[XY]$ should all be just numbers.


  • $X$ follows an exponential distribution, the expectation is $frac12$.


  • begin{align}
    E[XY] &= int_0^infty int_0^x 2y e^{-2x} , dy , dx \
    &=frac12int_0^infty 2x^2 e^{-2x}, dx \
    &= frac12 E[X^2]
    end{align}







share|cite|improve this answer









$endgroup$



Guide:




  • $E[X], E[Y], E[XY]$ should all be just numbers.


  • $X$ follows an exponential distribution, the expectation is $frac12$.


  • begin{align}
    E[XY] &= int_0^infty int_0^x 2y e^{-2x} , dy , dx \
    &=frac12int_0^infty 2x^2 e^{-2x}, dx \
    &= frac12 E[X^2]
    end{align}








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 3:09









Siong Thye GohSiong Thye Goh

102k1466118




102k1466118












  • $begingroup$
    I see. If i wanted to compute the integral $int x cdot f_{X}(x)$, what would the bounds be ?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:11










  • $begingroup$
    from $0$ to $infty$, that's where $f_X(x)$ is positive.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:12












  • $begingroup$
    ok let me try again
    $endgroup$
    – joseph
    Dec 13 '18 at 3:12










  • $begingroup$
    ok can you check again?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:20










  • $begingroup$
    seems fine to me.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:23


















  • $begingroup$
    I see. If i wanted to compute the integral $int x cdot f_{X}(x)$, what would the bounds be ?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:11










  • $begingroup$
    from $0$ to $infty$, that's where $f_X(x)$ is positive.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:12












  • $begingroup$
    ok let me try again
    $endgroup$
    – joseph
    Dec 13 '18 at 3:12










  • $begingroup$
    ok can you check again?
    $endgroup$
    – joseph
    Dec 13 '18 at 3:20










  • $begingroup$
    seems fine to me.
    $endgroup$
    – Siong Thye Goh
    Dec 13 '18 at 3:23
















$begingroup$
I see. If i wanted to compute the integral $int x cdot f_{X}(x)$, what would the bounds be ?
$endgroup$
– joseph
Dec 13 '18 at 3:11




$begingroup$
I see. If i wanted to compute the integral $int x cdot f_{X}(x)$, what would the bounds be ?
$endgroup$
– joseph
Dec 13 '18 at 3:11












$begingroup$
from $0$ to $infty$, that's where $f_X(x)$ is positive.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 3:12






$begingroup$
from $0$ to $infty$, that's where $f_X(x)$ is positive.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 3:12














$begingroup$
ok let me try again
$endgroup$
– joseph
Dec 13 '18 at 3:12




$begingroup$
ok let me try again
$endgroup$
– joseph
Dec 13 '18 at 3:12












$begingroup$
ok can you check again?
$endgroup$
– joseph
Dec 13 '18 at 3:20




$begingroup$
ok can you check again?
$endgroup$
– joseph
Dec 13 '18 at 3:20












$begingroup$
seems fine to me.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 3:23




$begingroup$
seems fine to me.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 3:23


















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