Computation of $P(3X+2Y < Z-W)$ when $W,X,Y,Z$ are independent normal variables

Question

If you are given this following problem.

$W,X,Y,Z$ are independent random variables with mean $= 1$ and standard deviation also equal to $1$. How would you compute this particular value: $P(3X+2Y < Z-W)$?

Integration is easy, but I am having trouble setting up the computation.

Answer 1

Don't approach this by integration. Since $W,X,Y,Z$ are iid $N(1,1^2)$ distributed, $3X+2Y-Z+W$ is $N(3+2-1+1, 3^2 + 2^2 + 1^2 + 1^2)=N(5,15)$ distributed. So, you only need to find the probability that a $N(5,15)$ random variable is negative which is standard.

Answer 2

Let $A=3X+2Y$ and $B=Z-W$

By linearity of expectation,

$$E(A)=E(3X+2Y)=E(3X)+E(2Y)=3E(X)+2E(Y)=5$$

By properties of variance $$Var(A)=Var(3X+2Y)=3^2 Var(X)+2^2 Var(Y)=13$$

We thus get

$$A\sim N(5, 13)$$

Similarly,

$$B\sim N(0,2)$$

Now let $C=A-B$ where now

$$C\sim N(5,15)$$

Then finally

$$\begin{align*} P(A\lt B) &=P(A-B<0)\\\\ &=P(C\lt 0)\\\\ &=\Phi\left(\frac{0-5}{\sqrt{15}}\right)\\\\ &\approx 0.098 \end{align*}$$

> pnorm(-5/sqrt(15))
[1] 0.0983528

R Simulation:

> w<-rnorm(10^6,1,1)
> x<-rnorm(10^6,1,1)
> y<-rnorm(10^6,1,1)
> z<-rnorm(10^6,1,1)
> mean(3*x+2*y<z-w)
[1] 0.098507

so our empirical result agrees with out analytical result fairly accurately.