Can someone please help me with finding the joint density of $Y_1$ and $Y_2$ when $Y_1=X_1+X_2$ and $Y_2=X_1-X_2$?
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1The general procedure is illustrated in the answer at http://stats.stackexchange.com/questions/32620/dividing-a-uniform-by-a-normal-random-variable-whats-the-distribution. – whuber Feb 14 '14 at 18:38
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Do you know anything about the distributions of X1 & X2? – gung - Reinstate Monica Feb 14 '14 at 18:51
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Hint: $\hat{Y}_1 = \displaystyle \frac{X_1+X_2}{\sqrt{2}}$ and $\hat{Y}_2 = \displaystyle \frac{X_1-X_2}{\sqrt{2}}$ are the coordinates of the random point $(X_1,X_2)$ measured with respect to axes that are rotated through an angle of $\pi/4$, and so it must be that $f_{\hat{Y}_1,\hat{Y}_2}(y_1, y_2)$ is just the joint density $f_{X_1,X_2}(x_1,x_2)$ with respect to the rotated axes. Now scale the random variables $\hat{Y}_1$ and $\hat{Y}_2$ by a factor of $\sqrt{2}$ to get the joint density of $Y_1$ and $Y_2$.
Dilip Sarwate
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Hint: It may also be worth noting the relationship between joint characteristic functions:
$\phi_{Y_1,Y_2}\left(t_1,t_2 \right)= \mathbb{E}\left[\mathrm e^{\mathrm i\left(t_1Y_1+t_2Y_2\right)} \right]=\mathbb{E}\left[\mathrm e^{\mathrm i\left[\left(t_1+t_2\right)X_1+\left(t_1-t_2\right)X_2 \right] }\right] = \phi_{X_1,X_2}\left(t_1+t_2,t_1-t_2 \right).$
Patrick Coulombe
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ir7
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