TITLE: Characteristic function of Normal random variable squared
QUESTION [5 upvotes]: Probability density function of $X^2$ when $X$ has $N(0,1)$ distribution
While reviewing above, 
Why do you sub $X^2$ for the $Y$ in $e^{tY}$ and not the density of the normal
$(\frac{1}{\sqrt{2\pi}}))e^{-Y^2 /2}?$
Also, could you find the $(U^2)$'s characteristic function?

REPLY [12 votes]: Direct evaluation of the characteristic function for $X \sim N(0,1)$
Observing that the expected value of a measurable function of $X$, $g(X)$, given that $X$ has a probability density function $f_X(x)$, is given by the inner product of $f_X$ and $g$:
$$
    \Bbb{E}(g(X)) = \int_{-\infty}^\infty g(x) f_X(x)\, \operatorname{d}x . 
$$
evaluating for $X\sim N(0,1)$ and $g(x)=e^{itx^2}$
$$
\Bbb{E}\left(e^{itX^2}\right) = \int_{-\infty}^\infty e^{itx^2} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\, \operatorname{d}x=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-ax^2}\, \operatorname{d}x=\frac{1}{\sqrt{2\pi}}\cdot\sqrt{\frac{\pi}{a}}
$$
where $a=\frac{1}{2}-it=\frac{1-2it}{2}$ and finally
$$
\varphi_{X^2}(t)=\Bbb{E}\left(e^{itX^2}\right) =(1−2it)^{−1/2}.
$$
Derivation of the pdf for $X \sim N(0,1)$
Let random variable $Y$ be defined as $Y = X^2$ where X has normal distribution with mean 0 and variance 1 (that is $X \sim N(0,1)$).
Then, for $y<0, \,\Bbb{P}(Y<y) = 0$ and for $y\ge 0$
$$
\begin{align}
P(Y<y) &= P(X^2<y) = P(|X|<\sqrt{y}) \\
& = F_X(\sqrt{y})-F_X(-\sqrt{y})\\
&= F_X(\sqrt{y})-(1-F_X(\sqrt{y}))\\
&= 2 F_X(\sqrt{y})-1 
\end{align}
$$
where $$ F_X\left(\sqrt{y}\right)= \int_{-\infty}^\sqrt{y} \frac{1}{\sqrt{2\pi}} \operatorname{e}^{-\frac{t^2}{2}} \operatorname{d}t$$
and the pdf is 
$$f_Y(y) = \frac{\operatorname{d}F_Y(y}{\operatorname{d} y}=2\frac{\operatorname{d}F_X(\sqrt{y})}{\operatorname{d} y} = 2 \frac{1}{\sqrt{2}\sqrt{\pi}} e^{-\frac{y}{2}} \left( \frac{1}{2} y^{-\frac{1}{2}} \right) = \frac{1}{2^{\frac{1}{2}} \Gamma(\frac{1}{2})}y^{\frac{1}{2}-1}e^{-\frac{y}{2}}  \Bbb{I}_{[0,+\infty)}(y) 
$$
Then $Y = X^2 \sim \chi^2(1)$.
Derivation of the characteristic function for $X \sim N(0,1)$
We now can find the characteristic function $\varphi_Y(t)$ as 
$$
\begin{align}
\varphi_Y(t)=\Bbb{E}\left(\operatorname{e}^{itY}\right)
&=\frac{1}{\sqrt{2\pi}}\int_0^\infty \frac{1}{\sqrt y}e^{-\frac{y}{2}+ity}\operatorname{d}y
=\frac{1}{\sqrt{2\pi}}\int_0^\infty \frac{1}{\sqrt y}e^{-y\left(\frac{1}{2}-it\right)}\operatorname{d}y
\end{align}
$$
and with the substitution $\sqrt y=u,\,\operatorname{d}y=2u\operatorname{d}u$ and $a=\frac{1}{2}-it=\frac{1-2it}{2}$ we have
$$
\varphi_Y(t)=\frac{1}{\sqrt{2\pi}}\int_0^\infty \frac{1}{u}e^{-au^2}2u\operatorname{d}u=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-au^2}\operatorname{d}u=\frac{1}{\sqrt{2\pi}}\cdot\sqrt{\frac{\pi}{a}}=\frac{1}{(1-2it)^{1/2}}
$$
that is 
$$
\varphi_Y(t)={(1-2it)^{-1/2}}
$$
Derivation for $X\sim N(\mu,\sigma^2)$
With direct calculation, obdserving that if $X\sim N(\mu,\sigma^2)$ then $X=\sigma Z+\mu$ for $Z\sim N(0,1)$, you'll have
$$
\varphi_{X^2}(t)=\varphi_{(\sigma Z+\mu)^2}(t)=\int_{-\infty}^{\infty}\operatorname{e}^{it(\sigma Z+\mu)^2}\frac{1}{\sqrt{2\pi}}\operatorname{e}^{-z^2/2}\operatorname{d}z
$$
Now putting $a=\frac{1}{2}-it\sigma^2=\frac{1-2it\sigma^2}{2}$ and $b=2it\mu\sigma$, and observing that
$$
\int_{-\infty}^{\infty}\operatorname{e}^{-a\xi^2+b\xi+c}\operatorname{d}\xi=\sqrt{\frac{\pi}{a}}\operatorname{e}^{\frac{b^2}{4a}+c}, \qquad(\mathbf{\text{prove it as exercise!}})
$$
you will find after some calculations
$$
\varphi_{X^2}(t)=\varphi_{(\sigma Z+\mu)^2}(t)=\frac{1}{(1-2i\sigma^2 t)^{1/2}}\exp\left(\frac{it\mu^2}{1-2i\sigma^2 t}\right)
$$
In similar way as before you'll find for $Y=X^2$ the pdf
$$
f_Y(y)=\frac{1}{\sigma\sqrt{2\pi y}}\operatorname{e}^{-\frac{(\sqrt y-\mu)^2}{2\sigma^2}} \Bbb{I}_{[0,+\infty)}(y) 
$$
and the characteristic function
$$
\varphi_Y(t)=\frac{1}{(1-2i\sigma^2 t)^{1/2}}\exp\left(\frac{it\mu^2}{1-2i\sigma^2 t}\right)
$$
Evaluation of the characteristic function for $U^2$
Let $X_1,\, \ldots,\,X_k$ be IID and $N(0,1)$; let $U^2$ be
$$
U^2=\sum_{j=1}^k X_j^2.
$$
Observing that each $X_j^2$ has characteristic function $\varphi_{X_j}(t)=(1−2it)^{−1/2}$ and that $X_1^2,\, \ldots,\,X_k^2$ are IID, then the characteristic function of $U^2$ is 
$$
\varphi_{U^2}(t)=\prod_{j=1}^k \varphi_{X_j}(t)=\prod_{j=1}^k (1−2it)^{−1/2}=(1−2it)^{−k/2}.
$$
If $X_1,\, \ldots,\,X_k$ be IID and $N(\mu,\sigma^2)$, observing that each $X_j^2$ has characteristic function $\varphi_{X_j}(t)=\frac{1}{(1-2i\sigma^2 t)^{1/2}}\exp\left(\frac{it\mu^2}{1-2i\sigma^2 t}\right)$ and that $X_1^2,\, \ldots,\,X_k^2$ are IID, then the characteristic function of $U^2$ is 
$$
\varphi_{U^2}(t)=\prod_{j=1}^k \varphi_{X_j}(t)=\prod_{j=1}^k \frac{1}{(1-2i\sigma^2 t)^{1/2}}\exp\left(\frac{it\mu^2}{1-2i\sigma^2 t}\right)=\frac{1}{(1-2i\sigma^2 t)^{k/2}}\exp\left(\frac{itk\mu^2}{1-2i\sigma^2 t}\right)
$$
{"set_name": "stack_exchange", "score": 5, "question_id": 597321}