2
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • 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, 2013 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, 2013 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, 2013 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, 2013 at 20:31

1 Answer 1

Reset to default
5
$\begingroup$

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
Improve this answer
$\endgroup$
2
  • $\begingroup$ This is the question I was about to ask. Could you briefly explain the math behind it? $\endgroup$
    – user51966
    Dec 10, 2018 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, 2018 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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