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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?

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  • $\begingroup$ The distribution of square of standard normal is chi-square. The sum of independent chi-square is still chi-square. $\endgroup$ – BGM Oct 24 '18 at 3:39
  • $\begingroup$ Can I find the distribution of one random variable since it is just chi square? $\endgroup$ – Anne Oct 24 '18 at 3:43
  • $\begingroup$ 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? $\endgroup$ – Anne Oct 24 '18 at 4:12
  • $\begingroup$ 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? $\endgroup$ – BGM Oct 24 '18 at 4:19
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    $\begingroup$ 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. $\endgroup$ – BGM Oct 24 '18 at 6:27
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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.

enter image description here

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
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  • $\begingroup$ 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 $\sigma^2.$ $\endgroup$ – BruceET Oct 24 '18 at 17:27

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