Derive chi-squared distribution from standard normal distribution

Question

I am supposed to derive chi-squared probability density function from standard normal distribution. However I am stuck with the first step - to calculate probability density function of $X^2$ where $$X - Normal(0,1).$$

I am using transformation of probability density function theorem: Let us have $X$ an n-dimensional random vector with probability density function $f_X$. Let us have $S_X$ open set such that $P(X \in S_X)=1$ and let $g: S_X \space->R^n$ be a diffeomorphism. Then for $Y=g(X)$ the probability density function is equal to $$f_Y(y)=f_X(g^{-1}(y)|J_{g^{-1}}(y)| \space for \space y \in g(S_X)$$

I have defined $g: u \space -> x^2$, calculated that $|J_{g^{-1}}(y))|=\frac{1}{2\sqrt{u}}$ and $g^{-1}:x\space ->\sqrt{u}$ and came to conclusion that $$f_U(u)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{u}{2}}\frac{1}{2\sqrt{u}}$$ for corresponding $u>0$. However this is not correct according to chi-squared distribution with 1 degree of freedom. There is one extra $\frac{1}{2}$. Am I missing something important? $$ $$ After this step I want to use convolution however I cannot go on without correct probability density function.

Answer 1

The formula you are using, referred to as "Function of random variables and change of variables in the probability density function" for scalar to scalar on Wikipedia, assumes monotonicity. So for X> 0 your function is monotonic and you can apply your theorem. Same goes for X <0. Then $F_y(y) = F_x(\sqrt{y}) -F_x(-\sqrt{y}) = F_x(\sqrt{y}) - (1- F_x(\sqrt{y})) = 2F_x(\sqrt{y})-1$. When you differentiate both sides the -1 becomes irrelevant. And there is your factor of 2.

For more detail see https://en.wikipedia.org/wiki/Proofs_related_to_chi-squared_distribution (Derivation of the pdf for one degree of freedom)