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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.

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    $\begingroup$ 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. $\endgroup$ 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
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    $\begingroup$ Yes, I think so. As you said: the CDF has no closed form (it cannot be expressed in term of elementary functions). $\endgroup$ Sep 7 '13 at 20:04
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    $\begingroup$ +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
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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
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  • $\begingroup$ 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
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    $\begingroup$ 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

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