Moment generating function of a binary variable
$begingroup$
We have a set of Random Variables $Y_i$ which takes the value $alpha$ with probability $(1-p)$ and takes the value $1-alpha$ with a probability of $p$.
We have been tasked with finding the Moment Generating Function (and the Cumulant Generating Function) of the sample mean of $Y_i$ (ie $S_N=frac{1}{N}sum^{N}_{i=0}Y_i$)
Knowing that I can simply use the formula:
$M_{S_N}(t)=[M_{Y_i}(frac{t}{N})]^N$
I just need to find the MGF of $Y_i$. However, I am having issues obtaining it's pdf and trying to get it into a nice form. I have tried defining it as:
$P(Y_i=y)=(1-p)delta_{y,alpha} + pdelta_{y,(1-alpha)}$
But still can't get the pdf into a usable form.
statistics binomial-coefficients moment-generating-functions
asked Dec 11 '18 at 2:04
ASRASR
11
$endgroup$
add a comment |
$begingroup$
We have a set of Random Variables $Y_i$ which takes the value $alpha$ with probability $(1-p)$ and takes the value $1-alpha$ with a probability of $p$.
We have been tasked with finding the Moment Generating Function (and the Cumulant Generating Function) of the sample mean of $Y_i$ (ie $S_N=frac{1}{N}sum^{N}_{i=0}Y_i$)
Knowing that I can simply use the formula:
$M_{S_N}(t)=[M_{Y_i}(frac{t}{N})]^N$
I just need to find the MGF of $Y_i$. However, I am having issues obtaining it's pdf and trying to get it into a nice form. I have tried defining it as:
$P(Y_i=y)=(1-p)delta_{y,alpha} + pdelta_{y,(1-alpha)}$
But still can't get the pdf into a usable form.
statistics binomial-coefficients moment-generating-functions
asked Dec 11 '18 at 2:04
ASRASR
11
$endgroup$
add a comment |
$begingroup$
We have a set of Random Variables $Y_i$ which takes the value $alpha$ with probability $(1-p)$ and takes the value $1-alpha$ with a probability of $p$.
We have been tasked with finding the Moment Generating Function (and the Cumulant Generating Function) of the sample mean of $Y_i$ (ie $S_N=frac{1}{N}sum^{N}_{i=0}Y_i$)
Knowing that I can simply use the formula:
$M_{S_N}(t)=[M_{Y_i}(frac{t}{N})]^N$
I just need to find the MGF of $Y_i$. However, I am having issues obtaining it's pdf and trying to get it into a nice form. I have tried defining it as:
$P(Y_i=y)=(1-p)delta_{y,alpha} + pdelta_{y,(1-alpha)}$
But still can't get the pdf into a usable form.
statistics binomial-coefficients moment-generating-functions
asked Dec 11 '18 at 2:04
ASRASR
11
$endgroup$
We have a set of Random Variables $Y_i$ which takes the value $alpha$ with probability $(1-p)$ and takes the value $1-alpha$ with a probability of $p$.
We have been tasked with finding the Moment Generating Function (and the Cumulant Generating Function) of the sample mean of $Y_i$ (ie $S_N=frac{1}{N}sum^{N}_{i=0}Y_i$)
Knowing that I can simply use the formula:
$M_{S_N}(t)=[M_{Y_i}(frac{t}{N})]^N$
I just need to find the MGF of $Y_i$. However, I am having issues obtaining it's pdf and trying to get it into a nice form. I have tried defining it as:
$P(Y_i=y)=(1-p)delta_{y,alpha} + pdelta_{y,(1-alpha)}$
But still can't get the pdf into a usable form.
statistics binomial-coefficients moment-generating-functions
statistics binomial-coefficients moment-generating-functions
asked Dec 11 '18 at 2:04
ASRASR
11
asked Dec 11 '18 at 2:04
ASRASR
11
asked Dec 11 '18 at 2:04
ASRASR
11
asked Dec 11 '18 at 2:04
ASRASR
11
asked Dec 11 '18 at 2:04
ASRASR
11
11
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
Why is this difficult? You simply apply the definition:
$$M_{Y_i}(t) = operatorname{E}[e^{tY_i}] = e^{talpha} Pr[Y_i = alpha] + e^{t(1-alpha)}Pr[Y_i = 1-alpha] = e^{talpha} (1-p) + e^{t(1-alpha)}p.$$
answered Dec 11 '18 at 2:08
heropupheropup
63.9k762102
$endgroup$
$begingroup$
So I reached this answer too, however my problem actually arose when obtaining the Cumulant Generating Function of $S_N$. My teacher has told me I should not be left with any logs when I simplify the equation, but with this MGF it seems impossible- so I thought the MGF might be wrong. I am pretty sure the teacher is just messing with me but just wanted to check (I really should have been clearer in my question, sorry)
$endgroup$
– ASR
Dec 11 '18 at 2:19
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Why is this difficult? You simply apply the definition:
$$M_{Y_i}(t) = operatorname{E}[e^{tY_i}] = e^{talpha} Pr[Y_i = alpha] + e^{t(1-alpha)}Pr[Y_i = 1-alpha] = e^{talpha} (1-p) + e^{t(1-alpha)}p.$$
answered Dec 11 '18 at 2:08
heropupheropup
63.9k762102
$endgroup$
$begingroup$
So I reached this answer too, however my problem actually arose when obtaining the Cumulant Generating Function of $S_N$. My teacher has told me I should not be left with any logs when I simplify the equation, but with this MGF it seems impossible- so I thought the MGF might be wrong. I am pretty sure the teacher is just messing with me but just wanted to check (I really should have been clearer in my question, sorry)
$endgroup$
– ASR
Dec 11 '18 at 2:19
add a comment |
$begingroup$
Why is this difficult? You simply apply the definition:
$$M_{Y_i}(t) = operatorname{E}[e^{tY_i}] = e^{talpha} Pr[Y_i = alpha] + e^{t(1-alpha)}Pr[Y_i = 1-alpha] = e^{talpha} (1-p) + e^{t(1-alpha)}p.$$
answered Dec 11 '18 at 2:08
heropupheropup
63.9k762102
$endgroup$
$begingroup$
So I reached this answer too, however my problem actually arose when obtaining the Cumulant Generating Function of $S_N$. My teacher has told me I should not be left with any logs when I simplify the equation, but with this MGF it seems impossible- so I thought the MGF might be wrong. I am pretty sure the teacher is just messing with me but just wanted to check (I really should have been clearer in my question, sorry)
$endgroup$
– ASR
Dec 11 '18 at 2:19
add a comment |
$begingroup$
Why is this difficult? You simply apply the definition:
$$M_{Y_i}(t) = operatorname{E}[e^{tY_i}] = e^{talpha} Pr[Y_i = alpha] + e^{t(1-alpha)}Pr[Y_i = 1-alpha] = e^{talpha} (1-p) + e^{t(1-alpha)}p.$$
answered Dec 11 '18 at 2:08
heropupheropup
63.9k762102
$endgroup$
Why is this difficult? You simply apply the definition:
$$M_{Y_i}(t) = operatorname{E}[e^{tY_i}] = e^{talpha} Pr[Y_i = alpha] + e^{t(1-alpha)}Pr[Y_i = 1-alpha] = e^{talpha} (1-p) + e^{t(1-alpha)}p.$$
answered Dec 11 '18 at 2:08
heropupheropup
63.9k762102
edited Dec 11 '18 at 2:51
answered Dec 11 '18 at 2:08
heropupheropup
63.9k762102
answered Dec 11 '18 at 2:08
heropupheropup
63.9k762102
answered Dec 11 '18 at 2:08
heropupheropup
63.9k762102
63.9k762102
$begingroup$
So I reached this answer too, however my problem actually arose when obtaining the Cumulant Generating Function of $S_N$. My teacher has told me I should not be left with any logs when I simplify the equation, but with this MGF it seems impossible- so I thought the MGF might be wrong. I am pretty sure the teacher is just messing with me but just wanted to check (I really should have been clearer in my question, sorry)
$endgroup$
– ASR
Dec 11 '18 at 2:19
add a comment |
$begingroup$
So I reached this answer too, however my problem actually arose when obtaining the Cumulant Generating Function of $S_N$. My teacher has told me I should not be left with any logs when I simplify the equation, but with this MGF it seems impossible- so I thought the MGF might be wrong. I am pretty sure the teacher is just messing with me but just wanted to check (I really should have been clearer in my question, sorry)
$endgroup$
– ASR
Dec 11 '18 at 2:19
$begingroup$
So I reached this answer too, however my problem actually arose when obtaining the Cumulant Generating Function of $S_N$. My teacher has told me I should not be left with any logs when I simplify the equation, but with this MGF it seems impossible- so I thought the MGF might be wrong. I am pretty sure the teacher is just messing with me but just wanted to check (I really should have been clearer in my question, sorry)
$endgroup$
– ASR
Dec 11 '18 at 2:19
$begingroup$
So I reached this answer too, however my problem actually arose when obtaining the Cumulant Generating Function of $S_N$. My teacher has told me I should not be left with any logs when I simplify the equation, but with this MGF it seems impossible- so I thought the MGF might be wrong. I am pretty sure the teacher is just messing with me but just wanted to check (I really should have been clearer in my question, sorry)
$endgroup$
– ASR
Dec 11 '18 at 2:19
add a comment |
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