Normal distribution squared probability

Question

Let $X_1,X_2,X_3,X_4$ be independent standard normal random variables and $Y=X^2_1+X^2_2+X^2_3+X^2_4$. Find the probability that $Y≤3$. Enter your answer as a decimal and make sure that at least $10$ digits after the decimal point are correct.

I am confused about something. Since the normal distribution of $X$ is ${1 \over {\sqrt{2\pi}}}\times e$ should just square that for $X_1$, $X_2$, $X_3$ and $X_4$? And then add them up? That way I will have four variables. Any intuition as to how to proceed will be appreciated.

Answer 1

Your conditions imply that $Y$ has a $\chi_4^2$ distribution. So you can just look up the answer in standard chi-square tables.

Answer 2

As others have noted the sum of the squares of standard normal random variables is distributed as a chi-squared distribution. In this case, a sum of length 4 means it's a chi-squared with four degrees of freedom. $$ Y\sim \chi_4^2 $$

The probability density function for a chi-squared distribution with $k$ degrees of freedom is $$ f_Y(y) = \frac{1}{2^{k/2}\Gamma(k/2)} y^{k/2-1} e^{-y/2}, $$ where $\Gamma$ is the gamma function that generalizes the factorial. In your case when $k=4$, note that $\Gamma(2)=(2-1)!=1$, so this reduces to $$ f_Y(y) = \frac{1}{4} y e^{-y/2} $$ Now the question is what is the probability that $0<Y<3$. So we integrate the density: $$ \Pr[0<Y<3] = \int_0^3 f_Y(y) dy = \int_0^3 \frac{1}{4} y e^{-y/2} dy $$ You can integrate this to get $$ \int_0^3 \frac{1}{4} y e^{-y/2} dy = 1-\frac{5}{2}e^{-3/2} \approx 0.442174599629, $$ or as a percentage with the requested number of digits after the decimal $$ \Pr[0<Y<3] = 44.2174599629\% $$