Since the distribution of a difference of two normally distributed variates X and Y with means and variances $(\mu_x,\sigma_x^2)$ and $(\mu_y,\sigma_y^2)$ respectively is given by another normal distribution with mean $\mu_x-\mu_y$ and variance $\sigma_x^2+\sigma_y^2$ http://mathworld.wolfram.com/NormalDifferenceDistribution.html.

Assuming that the distributions are independent, can we find a distribution of a difference of two multivariate normally distributed variates?


1 Answer 1


If two multivariate normal random variables $X\sim\mathcal N(\mu_X,\Sigma_X)$ and $Y\sim\mathcal N(\mu_Y,\Sigma_Y)$ are independent (and of the same dimension), then their sum is still normal, and you can sum the mean and variance directly: $$X+Y\sim\mathcal N(\mu_X+\mu_Y,\Sigma_X+\Sigma_Y).$$

The easiest way to see this is with the expression of the characteristic function, as can be found on the Wikipedia article for example: $$\phi_X(u)=\exp(iu'\mu_X-u'\Sigma_Xu).$$

Since the characteristic function of the sum of two independent random variables is the product of their respective characteristic functions, then \begin{eqnarray} \phi_{X+Y}(u) &=& \exp(iu'\mu_X-u'\Sigma_Xu)\exp(iu'\mu_Y-u'\Sigma_Yu)\\ &=& \exp(iu'\mu_X-u'\Sigma_Xu+iu'\mu_Y-u'\Sigma_Yu)\\ &=& \exp(iu'(\mu_X+\mu_Y)-u'(\Sigma_X+\Sigma_Y)u). \end{eqnarray}

Now back to your original question, it suffices to see that $-Y$ has the same variance as $Y$, but with opposite mean to conclude: $$-Y\sim \mathcal N(-\mu_Y,\Sigma_Y),$$ so indeed $$X-Y\sim\mathcal N(\mu_X-\mu_Y,\Sigma_X+\Sigma_Y).$$

  • $\begingroup$ Is this true for the case that one variable is bivariate normal and the other is univariate normal? $\endgroup$
    – kathystehl
    Dec 17, 2015 at 18:49
  • $\begingroup$ Oh... characteristic function.. not an easy concept to grasp. Would you have a proof without it? $\endgroup$
    – Tomas
    Oct 28, 2019 at 18:09

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