# Finding the pdf of the squared difference between two independent standard normal random variables

Let $X_1$ and $X_2$ be independent $N(0,1)$ random variables. Find the pdf of $(X_1-X_2)^2/2$.

I know that the answer will be a chi-squared with 1 degree of freedom. But I am not sure how to show this formally. Any suggestions are greatly appreciated.

## 1 Answer

Start by showing that the distribution of $(X_1 - X_2)$ is normal - Distribution of the difference of two normal random variables.

Then use the definition of the chi-squared distribution.

EDIT:
If you have seen the above mentioned question, you would have concluded that $X_1 - X_2$ follows a normal distribution with mean $0$ and variance $2$.
Hence, the R.V. - $\frac{(X_1-X_2)}{\sqrt{2}}$ follows a standard normal distribution. This is because $$\operatorname {Var} (aX)=a^{2}\operatorname {Var} (X).$$ Using the definition as stated here, conclude that the square of a standard normal random variable follows the chi-squared distribution with $k=1$ degrees of freedom.

• I did the first part of your suggestion but I do not see how using the definition of chi-square distribution will help me there after, could you add a bit more to your answer – Wolfy Nov 25 '16 at 17:38
• @Wolfy - Explained a bit more, hope this clarifies. – Shraddheya Shendre Nov 25 '16 at 17:56
• the post you have contains the wrong answer – Wolfy Dec 4 '16 at 18:41
• You mean the linked answer? – Shraddheya Shendre Dec 5 '16 at 4:59