# Characteristic function of Normal random variable squared

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?

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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 $$\varphi_{U^2}(t)=\prod_{j=1}^k \varphi_{X_j}(t)=\prod_{j=1}^k \frac{1}{(1-2i\sigma^2 t)^{1/2}}\exp\left(\frac{it\mu^2}{1-2i\sigma^2 t}\right)=\frac{1}{(1-2i\sigma^2 t)^{k/2}}\exp\left(\frac{itk\mu^2}{1-2i\sigma^2 t}\right)$$

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but, would you sub X^2 into the density also? – Joey Dec 8 '13 at 0:32
So you would have (1/(sqrt(2*pi*var)))*(e^-(((x^2)-u)^2)/(2(var)))??? – Joey Dec 8 '13 at 0:35
*for a non standard normal – Joey Dec 8 '13 at 0:37
I edit a bit more... – alexjo Dec 8 '13 at 13:09