Let $(X,Y)\sim N(\mu_x=1,\mu_y=1,\sigma^2_x=4,\sigma^2_y=1,\rho=1/2)$. Compute $P(X+2Y\leq 4)$.

How do you compute probabilities of a bivariate normal? For a regular normal distribution I remember we had to use tables or software because there is no close form solution for the normal CDF.

  • 6
    $\begingroup$ I think you don't even need CDF of the bivariate normal distribution. You can determine the mean and variance of the distribution $Z: X+2Y$ and then use the CDF of a univariate normal distribution. $\endgroup$ – COOLSerdash Sep 7 '13 at 19:58
  • $\begingroup$ By "use the CDF of a nunivarate normal" you mean look up the value in a table or use software? $\endgroup$ – bdeonovic Sep 7 '13 at 20:02
  • 1
    $\begingroup$ Yes, I think so. As you said: the CDF has no closed form (it cannot be expressed in term of elementary functions). $\endgroup$ – COOLSerdash Sep 7 '13 at 20:04
  • 5
    $\begingroup$ +1 to @COOLSerdash. It's worth saying explicitly that one characterization of the multivariate normal distribution is that every linear combination of the elements of a multivariate normal vector is normally distributed. This is why calculating the mean and variance of $X+2Y$ is sufficient to answer the question. $\endgroup$ – Macro Sep 7 '13 at 20:31

Since $(X,Y)$ is normally distributed then $X+2Y\sim N(\mu_x+2\mu_y, \sigma_x^2+4\rho \sigma_x\sigma_y+4\sigma_y^2)\sim N(3, 12)$. Using R we get

> pnorm(4,mean=3,sd=sqrt(12))
[1] 0.613585
| cite | improve this answer | |
  • $\begingroup$ This is the question I was about to ask. Could you briefly explain the math behind it? $\endgroup$ – user51966 Dec 10 '18 at 14:52
  • 1
    $\begingroup$ Normal distribution is closed under affine transforms (If $X\sim N(\mu, \Sigma)$ then $\mathbf{c}+\mathbf{A}X \sim N(c+\mathbf{A}\mu, \mathbf{A}\Sigma\mathbf{A}^\intercal)$). $\endgroup$ – bdeonovic Dec 17 '18 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.