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Let $Y_1, Y_2,$ and $Y_3$ be independent, $\mathrm{N}(0,1)$-distributed random variables, and set $$\begin{cases}X_1 = Y_1-Y_3 \\ X_2 = 2Y_1 + Y_2 - 2Y_3 \\X_3 = -2Y_1 + 3Y_3 \end{cases}. $$ Determine the conditional distribution of $X_2$ given that $X_1+X_3 = x$.

So, what I need to compute is $$\frac{f_{X_2, X_1+X_3}(x_2,x_1+x_3)}{f_{X_1+X_3}(x_1+x_3)}. $$

First, I thought I could find the joint distribution $f_{X_2,X_1+X_3}$. When computing $\mathrm{Cov}(X_2,X_1+X_3)$, I find that it is equal to zero. Meaning that $X_2$ and $X_1+X_3$ are independent if they are jointly normal. But if they are not, then how do I compute their joint distribution?

If they were in fact independent, then it would be enough to compute $f_{X_2}(x_2)$ and with $X_2 = 2Y_1+Y_2-2Y_3$ that would have the distribution $\mathrm{N}(B\mu, B\Lambda B')$ with $B= [2, 1, -2]$, but that gives me the result $\mathrm{N}(0,9)$, which is wrong. So $X_2$ and $X_1+X_3$ can't be independent, so they can't be jointly normal, so I have no idea how to compute their joint distribution. I do know how to compute $f_{\bar{Y}}(\bar{y})$ and $f_{\bar{X}}(\bar{x})$, but other than that, I know nothing, it feels like.

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  • $\begingroup$ Have you double checked your covariance computation? $\endgroup$ – BGM Mar 6 '18 at 11:05
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So I actually - finally - figured it out, and just in case anyone else is ever stuck on this, I'll post what I did.

Throughout I've been using the following formula: That for a random vector $X \in \mathrm{N}(\mu,\Lambda),$ the random vector $Y = BX$ has distribution $\mathrm{N}(B\mu, B\Lambda B')$.

By setting $$B = \begin{pmatrix}1 & 0 &-1 \\2 & 1 & -2\\-2&0&3 \end{pmatrix}$$ I found that $f_{\bar{X}}(\bar{x}) \in \mathrm{N}(0, \Lambda_{\bar{X}}),$ with $$\Lambda_{\bar{X}} = B \underbrace{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1 \end{pmatrix}}_{\Lambda_{\bar{Y}}}B',$$ where the matrix $\Lambda_{\bar{Y}}$ is the covariance matrix for $(Y_1,Y_2,Y_3)'.$ I then did another similar transformation in order to find $f_{X_2, X_1+X_3}(x_2,x_1+x_3) = f_{X_2,U}(x_2,u)$ by setting $$\hat{B} = \begin{pmatrix}0&1&0\\1&0&1 \end{pmatrix}, $$ and this way finding $\Lambda_{(X_2,U)}.$ In other words, I computed $$f_{X_2,U}(x_2,u) = \frac{1}{\sqrt{2\pi}\sqrt{\mathrm{det}(\Lambda_{(X_2,U})}}\mathrm{exp}\left(-\frac{1}{2}\left((x_2,u)\Lambda_{(X_2,U)}^{-1}(x_2,u)' \right) \right). $$

I did the same for $f_{X_1+X_3}(x_1+x_3) = f_U(u)$ by using $\tilde{B} = (1,0,1)$, and then I could compute $$\frac{f_{X_2,U}(x_2,u)}{f_U(u)}. $$

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