Does $Y=\alpha X + \beta$ hold for multivariate gaussian density?

Question

In the one-dimensional case, if $X$ is $\mathcal{N}(\mu,\sigma^2)$, then $Y =\alpha X + \beta $ is $\mathcal{N}(\alpha \mu + \beta,\alpha^2\sigma^2)$ . We can prove this using the cumulative distribution function of of $Y$

$F_Y(a) = P\{Y \leq a\} = P\{\alpha X + \beta \leq a\} = P\{X \leq (a-\beta)/\alpha\}$.

Substituting $Y =\alpha X + \beta $ and change of variable gives us,

$F_Y(a) = \int_{-\infty}^{a} \frac{1}{\sqrt{2\pi}(\alpha\sigma)} \exp \{ \frac{-(v-(\alpha \mu + \beta))^2}{2(\alpha\sigma)^2}\} dv $

Hence $f_Y(v) = \frac{1}{\sqrt{2\pi}(\alpha\sigma)} \exp \{ \frac{-(v-(\alpha \mu + \beta))^2}{2(\alpha\sigma)^2}\} $

Thus $Y$ is $\mathcal{N}(\alpha \mu + \beta, \alpha^2 \sigma^2)$.

In the multivariate case, if $X$ is $\mathcal{N}(\mu,\Sigma)$ and $Y=\alpha X + \beta$, is $Y \sim \mathcal{N}(\alpha \mu + \beta,\alpha^2\Sigma)$? If so, how do we prove it?

Answer 1

The method of characteristic functions (CF) will work here. So we have the CF for $X$ as $$\varphi_{X}(t)=\exp\left(it^{T}\mu_{X}-\frac{1}{2}t^{T}\Sigma_{X}t\right)$$ Now we make the substitution $Y=\alpha X + \beta$ in the CF and we get:

$$\varphi_{Y}(t)=E\left[\exp(it^{T}Y)\right]=E\left[\exp(it^{T}\alpha X +it^{T}\beta)\right]=\exp(it^{T}\beta)\varphi_{X}(\alpha t)$$

Then substitute in the CF expression for $X$.

$$\varphi_{Y}(t)=\exp(it^{T}\beta)\exp\left(i(\alpha t)^{T}\mu_{X}-\frac{1}{2}(\alpha t)^{T}\Sigma_{X}(\alpha t)\right)$$ $$=\exp\left(it^{T}[\alpha\mu_{X}+\beta]-\frac{1}{2}t^{T}[\alpha^{2}\Sigma_{X}]t\right)$$

But this is the characteristic function of a new normal distribution with mean vector $\alpha\mu_{X}+\beta$ and covariance matrix $\alpha^{2}\Sigma_{X}$. As characteristic functions are uniquely defined from a distribution function and vice versa, you have your proof.

To generalise to the case where $\alpha$ is an appropriately defined $c\times p$ matrix ($p$ is the dimension of $X$). we simply replace the covariance matrix $\alpha^{2}\Sigma_{X}$ with the $c\times c$ covariance matrix $\alpha\Sigma_{X}\alpha^{T}$. Note that $\beta$ must be a $c\times 1$ vector for mean vector to make sense - but it is unchanged at $\alpha\mu_{X}+\beta$.