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04:29
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A: Evaluating the mean and variance of the square of a normal distribution

Siong Thye GohNot quite right. You did not address the crossed term. $Y=(X_1+2X_2+X_3)^2=X_1^2+4X_2^2+X_3^2+4X_1X_2+2X_1X_3+4X_2X_3$ Hence,by independenc and linearity of expectation $E[Y] = E[X_1^2]+4E[X_2^2]+E[X_3^2]+4E[X_1]E[X_2]+2E[X_1]E[X_3]+4E[X_2]E[X_3]$ Now, you can evaluate the individual terms. Now,...

don't you think cross-terms be zero as x1,x2,x3 are independent?
No. Independence just means $E[X_1X_2]=E[X_1]E[X_2]$.
got it, thanks. and to evaluate E(X1^2) I have to use variance formula, right?
yup. $E[X_1^2]=Var(X_1)+E[X_1]^2$.
I got E[Y] = 77 using the first approach. Could you please tell me how to use the second moment of the normal distribution to get this?
04:29
$E[Y]=E([(X_1+2X_2+X_3)^2]$ this is equal to the second moment of $X_1+2X_2+X_3$ which is just equal to $2^2+8^2+3^2=77$ since the expectation of $X_1+2X_2+X_3$ is $0$.
Thank you for this approach. One more doubt to compute 𝜎̃ ^4 I have to 2^4+8^4+3^4 or 77^2?
hi, since $\tilde{\sigma}^2 = 77$, we have $\tilde{\sigma}^4 = 77^2$.
in general, square of the sum is not equal to the sum of square of each term
just starting the chat to avoid long comment threads
Hello
so, i got E[Y^2] =12579
how do you obtain that number?
3(2^4+8^4+3^4) = 12579
04:35
we can't just power 4 each individual term
So, how to compute E[Y^2] using the 4th moment of norma distribution?
the formula is $3\tilde{\sigma}^4$
the formula for the moments can be found on the wikipedia page
how to compute sigma^4?
so, E[Y^2] will be 17787
3 * 77*77
04:41
yup
variance is coming 11858 ( 17787 - 77^2)
seems to be so
Thank you so much, Siong
welcome. remember for future post, always include your attempt and use descriptive title. :)
yes, I'll. I'm new to stackexchange still figuring out things but will surely do from next time
04:47
welcoem to the site, don't be discouraged from the negative votes :) see you around the site sometimes
Thank you so much
one more question, can you tell what distribution will be E[Y] having?
normal distribution, right?
 
11 hours later…
16:02
E[Y] is a number

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