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Random variables X and Y are independent identically distributed standard normal random variables. Consider random variables U=aX+bY and V=aX-bY, where a and b some fixed numbers and a^2 +b^2 =1.

a. Determine the joint pdf fuv, using the Jacobian formula.

b. Determine the conditional pdf fu|y(u|y=1).

I solved the part a by applying Jacobian's formula. I think the random variable should be normal distributed. However, after I simplify it, it looks very complicated.

I got determinant det(Ju,v(x,y))=2ab. fu,v=(fx,y(U+V/2a, U-V/2b))/|det(Ju,v)|. since X,Y are iid, (fx(U+V/2a)*fy(U-V/2b))/2ab=(1/4abPi)exp(-(u^2-2(a^2-b^2)uv+v^2)/8a^2b^2))

I donot know where I made mistakes. I also confused about part b. Could anyone help me out? Thank you very much.

Deepsea234
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  • Random variable X and Y are independent identically distributed standard normal random variable. Consider random variables U=αX+βY and V=αX-βY, where α,β are some fixed numbers and α^2+β^2=1. Determine the joint pdf f_(U,V) using Jacobian formula. Determine the conditional pdf f_(U|Y) (u│y=1). My answer to part a: U=(X+Y)/2α, V=(U-V)/2β. Det(J_(U,V))=2αβ. f(U,V)=(1/2αβ)(1/2π) e^(-((u+v)/2α)^2-((u-v)/2β)^2 ) = 1/4αβπ e(-(u^2-2(α^2-β^2 )uv+v^2 )/8α^2 β^2) – Deepsea234 Nov 03 '17 at 02:08

1 Answers1

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There is not one but two random variables $U$ and $V$ and they are jointly bivariate normal. Although you've written it out pretty unreadably and I don't feel like parsing it, it looks like you're on the right track at the very least. This joint PDF might look a little unfamiliar to you. You can find it here under the heading "Bivariate case" but you might have to do some algebra to get it into this form (You can compute the means, variances and correlation of $U$ and $V$ from standard formulae).

For the conditional $PDF,$ since $X$ and $Y$ are independent you can basically set $Y=1$ and write $U\mid_{Y=1} = aX +b$ so it has the same distribution as $aX+b.$ (If they weren't indpendent you'd need to use the conditional distribution of $X$ given $Y$.) Alternatively you could compute the joint distribution of $U$ and $Y$ by similar methods as above and then use the formula $$ f(u\mid y) = \frac{f(u,y)}{f(y)}.$$

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spaceisdarkgreen
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  • I think I confused about the symbol. The professor did not explicitly explain it as you did. But I still feel my joint density function was wrong although I was on the right track. If you have some time, would you show the solution? Thank you very much for your help. – Deepsea234 Nov 03 '17 at 03:24
  • I came up with different solutions for part b. Since X, Y are independent, I worked on the distribution aX+b. However, which function do I need to start with? pdf of fx or pdf of fu,v? Would you explain it further for me? Thank you. – Deepsea234 Nov 03 '17 at 14:57