Finding the Probability from the sum of 3 random variables

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?

• The distribution of square of standard normal is chi-square. The sum of independent chi-square is still chi-square. – BGM Oct 24 '18 at 3:39
• Can I find the distribution of one random variable since it is just chi square? – Anne Oct 24 '18 at 3:43
• I am still confused as to where I can apply this, since we know it is chi-square distribution could I find the distribution of Z random variable and find the pdf from that, then the probability with 3 degree of freedoms since there are 3 random variables? – Anne Oct 24 '18 at 4:12
• I see that you say chi square with $3$ degrees of freedom, so seems you somehow have the concept, but not sure which part you confused. I do not state all the things explicitly because that will be a spoiler. Can you state which variable has what distribution clearly and what confusion do you have now? – BGM Oct 24 '18 at 4:19
• Why you say $Z_1^2 + Z_2^2 + Z_3^2$ has the same distribution as $3Z^2$? Those three random variables are independent, they are not the same variable. As you said the sum is just a chi-squared random variable with 3 degrees of freedom. – BGM Oct 24 '18 at 6:27

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

• This problem relates to practical issues as well as learning about chi-sq dist'n: If data are normal, then the sample variance $S^2$ estimates the population variance $\sigma^2$ and the relationship $(n-1)S^2/\sigma^2 \sim$ CHISQ$(n-1)$ provides a confidence interval for $\sigma^2.$ – BruceET Oct 24 '18 at 17:27