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I have a multivariate normal distribution $Y \sim \mathcal{N}(\mu,\Sigma)$ and a non-singular matrix $B$. I'd like to find the distribution of $X = B \cdot Y$.

So far I've written \begin{align*} \mathbb{P}(X \leq a) &=\mathbb{P}(B \cdot Y \leq a) =\mathbb{P}(Y \leq B^{-1} a) = f_Y(B^{-1}a) \\[0.2cm] &= \frac{1}{\sqrt{(2 \pi)^n \det(\Sigma)}} \cdot \exp \Big( -\frac{1}{2} (B^{-1}a - \mu)^T \Sigma^{-1} (B^{-1}a - \mu) \Big). \end{align*}

I think that $X \sim \mathcal{N}(\mu', \Sigma)$, for some $\mu'$, but I have no idea how to show this.

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    $\begingroup$ $$B\cdot Y \sim \mathcal{N}(B\cdot \mu,B\cdot \Sigma\cdot B^T)$$ $\endgroup$ – Did Oct 12 '16 at 12:48
  • $\begingroup$ To me the following expression seems to be wrong:$$\mathbb{P}(B \cdot Y \leq a) = \mathbb{P}(Y \leq B^{-1} a) = f_Y(B^{-1}a)$$ because $$f_Y$$ looks like a pdf and $$\mathbb{P}(B \cdot Y \leq a) $$ looks like a cdf. $\endgroup$ – zoli Oct 12 '16 at 12:59
  • $\begingroup$ almost duplicate of (math.stackexchange.com/q/332441). $\endgroup$ – Jean Marie Oct 12 '16 at 13:08
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The wrong point in your reasoning is the equivalence:

$$\tag{1}BY \leq a \ \ \ \ \Leftrightarrow \ \ \ \ Y \leq B^{-1}a$$

Here is a simple counter example to (1) using $B=\pmatrix{1 & 0 \\ 0 & -1}=B^{-1}$. More precisely, if (1) were true, we would have the following equivalence:

$$\pmatrix{1 & 0 \\ 0 & -1}\pmatrix{Y_x\\Y_y} \leq \pmatrix{a_x\\a_y} \ \ \ \ \Leftrightarrow \ \ \ \ \pmatrix{Y_x\\Y_y}\leq \pmatrix{1 & 0 \\ 0 & -1}\pmatrix{a_x\\a_y}$$ Otherwise said:

$$\cases{\ \ Y_x \leq a_x\\-Y_y \leq a_y} \ \ \ \ \Leftrightarrow \ \ \ \ \cases{Y_x \leq \ \ a_x\\Y_y \leq -a_y}$$

The equivalence between the second lines is evidently false in general...

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