Probability that linear combination of normal random variables exceeds a value

I'm new to statistics, and I'm wondering how can I compute $$P(2Y_1 + 4Y_2 - 3Y_3 \geq 40)$$ given the following information?

• $$Y$$ follows a multivariate normal distribution.

• $$Y_1, Y_2, Y_3$$ follow a normal distribution.

• $$Y_1, Y_2, Y_3$$ have means $$10, 12, 14$$.

• $$Y_1, Y_2, Y_3$$ have variances $$2, 2, 8$$.

• The covariance between $$Y_1, Y_2$$ is $$0.50$$, and the covariance between $$Y_1, Y_3$$ is $$-0.75$$. The variables $$Y_2$$ and $$Y_3$$ are independent.

Here's the covariance matrix I found for $$Y_1, Y_2, Y_3$$:

$$\begin{bmatrix} 2 & .50 & -.75 \\ .5 & 2 & 0 \\ -.75 & 0 & 8 \end{bmatrix}$$

I know how to do it when $$Y_1, Y_2, Y_3$$ are all independent by using the fact that linear combinations of independent normal random variables are normal. However, I'm really not sure about how to do it in this case. I can easily find the covariance matrix, but I'm not sure how to proceed from there.

• Just knowing their means, variances and covariances along with the fact that each variables is (marginally) normal is insufficient. Sep 18, 2020 at 3:32

If you're looking for an analytic solution, this is what you should do. Start by defining a new variable $$X = 2Y_1 + 4Y_2 - 3Y_3$$ A sum of jointly normal random variables is also normal, even if the variables are not independent (see Wikipedia article here). All that remains is to compute the mean and variance.

The mean of $$X$$ can be determined as \begin{align*} E[X] &= 2 E[Y_1] + 4 E[Y_2] - 3 E[Y_3] \\ &= (2 \times 10) + (4 \times 12)-(3 \times 14) \\ &= 26 \end{align*} The variance of $$X$$ is slightly easier to write in matrix form. Let $$Y \equiv (Y_1,Y_2,Y_3)$$ and $$\omega \equiv (2,4,-3)$$ be vectors of random variables and weights, respectively. The variance is given by

$$Var(X) = Var(\omega'Y) = \omega'Var(Y)\omega$$

Numerically this is equal to $$Var(X) = \begin{bmatrix} 2 & 4 & -3 \end{bmatrix} \begin{bmatrix} 2& .5 & -.75 \\ .5 & 2 & 0 \\ -.75 & 0 & 8 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \\ -3 \end{bmatrix}= 129$$ If $$\Phi$$ is the CDF of a standard normal, \begin{align*} P(X > 40) &= 1-P\left( X \le 40 \right) \\ &= 1-P\left( \frac{X - E[X]}{\sqrt{Var(X)}} \le \frac{40 - E[X]}{\sqrt{Var(X)}} \right) \\ &= 1-P\left( Z \le \frac{40 - E[X]}{\sqrt{Var(X)}} \right) \\ &= 1-\Phi\left( \frac{40 - E[X]}{\sqrt{Var(X)}} \right) \\ &= 0.108867 \end{align*}

• "A sum of normal random variables is also normal, even if the variables are not independent" - I don't think that's true. Do you have a proof?
– Paul
Sep 17, 2020 at 7:31
• @Paul you are correct, the necessary assumption is that the random variables are jointly gaussian (which they are in this case). The easy counterexample is $X \sim \mathcal N (0,1)$ and $Y = -X$. Then $X + Y = 0$ is not gaussian. Sep 17, 2020 at 7:55
• Another nice counterexample is here math.stackexchange.com/questions/563364/… Sep 17, 2020 at 8:05
• Great counter examples! I have amended the statement to emphasize that "joint" normality is important. Your comments emphasize the pitfalls of focusing on the marginal distributions only. Sep 18, 2020 at 2:56

You can obtain the mean and variance of the resulting linear combination or you can also do a simulation and obtain the results.

Via simulation.

> covmat=matrix(c(2,0.5,-0.75,0.5,2,0,-0.75,0,8),nrow=3)
> is.positive.definite(x, tol=1e-8)
[1] TRUE
> means=c(10,12,14)
> weights=c(2,4,-3)
> mat=mvrnorm(10^7,means,covmat)
> mat=mat %*% weights
> result=sum(mat>40)/10^7
> result
[1] 0.1088587


Non simulated

The mean is a linear combination of means

> new_mean= means %*% weights
> new_mean
[1] 26


The variance is obtained multiplying

> weights %*% covmat %*% weights
[,1]
[1,]  129


> 1-pnorm(40,26,sqrt(129))