Finding the Probability from the sum of 3 random variables

Question

Let $X_1, X_2$ and $X_3$ be three independent normal random variables having mean $\mu= 0$ and variance $\sigma^2=16.$

Compute $P(X_1^2+X_2^2+X_3^2>8).$

Hint: First transform the random variables to standard normal.

I transformed the random variables to $Z$ standard normal and got $Z_1=X_1/4,\, Z_2=X_2/4$ and $Z_3=X_3/4.$ I am unsure about where to go from here.

I know that the sum of random variables is the same as the product of their moment generating functions but how do I apply that here?

Answer 1

Suggested outline.

(1) Use MGFs (or a transformation method) to show that each of the three $Z_i,$ for $i=1,2,3,$ has a chi-squared distribution with $1$ degree of freedom.

(2) Use MGFs to show that $Q = Z_1^2 + Z_2^2 + Z_3^2$ has a chi-squared distribution with $3$ degrees of freedom.

(3) Use software or printed tables of the distribution $\mathsf{Chisq}(df=3)$ to evaluate (or approximate) $P(Q > 0.5).$

Using R statistical software, I get about 0.9189. (Depending on the printed tables available, you may be able to say only that the answer is between .90 and .95.)

1-pchisq(.5, 3)
[1] 0.9188914

In the figure below, the desired probability is represented by the area under the density curve to the right of the vertical red line at 0.5.

Note: A simulation in R, accurate to two or three places.

set.seed(1024);  m = 10^6
x1 = rnorm(m, 0, 4)
x2 = rnorm(m, 0, 4)
x3 = rnorm(m, 0, 4)
s = x1^2 + x2^2 + x3^2
mean(s > 8)
## 0.918736