gamma distributionnormal distributionself-studyvariance
So I'm reading about the derivation of the variance for normal distribution and I don't understand the following derivation with the use of gamma function.
So, if I continue this derivation the integral becomes
$$
2\int_{-\infty}^\infty ue^{-u}du\
$$
which is clearly not gamma function (in gamma function integral goes from 0 to infinity). How can I solve this integral?
As Prof. Sarwate's comment noted, the relations between squared normal and chi-square are a very widely disseminated fact - as it should be also the fact that a chi-square is just a special case of the Gamma distribution:
$$X \sim N(0,\sigma^2) \Rightarrow X^2/\sigma^2 \sim \mathcal \chi^2_1 \Rightarrow X^2 \sim \sigma^2\mathcal \chi^2_1= \text{Gamma}\left(\frac 12, 2\sigma^2\right)$$
the last equality following from the scaling property of the Gamma.
As regards the relation with the exponential, to be accurate it is the sum of two squared zero-mean normals each scaled by the variance of the other, that leads to the Exponential distribution:
$$X_1 \sim N(0,\sigma^2_1),\;\; X_2 \sim N(0,\sigma^2_2) \Rightarrow \frac{X_1^2}{\sigma^2_1}+\frac{X_2^2}{\sigma^2_2} \sim \mathcal \chi^2_2 \Rightarrow \frac{\sigma^2_2X_1^2+ \sigma^2_1X_2^2}{\sigma^2_1\sigma^2_2} \sim \mathcal \chi^2_2$$
$$ \Rightarrow \sigma^2_2X_1^2+ \sigma^2_1X_2^2 \sim \sigma^2_1\sigma^2_2\mathcal \chi^2_2 = \text{Gamma}\left(1, 2\sigma^2_1\sigma^2_2\right) = \text{Exp}( {1\over {2\sigma^2_1\sigma^2_2}})$$
But the suspicion that there is "something special" or "deeper" in the sum of two squared zero mean normals that "makes them a good model for waiting time" is unfounded: First of all, what is special about the Exponential distribution that makes it a good model for "waiting time"? Memorylessness of course, but is there something "deeper" here, or just the simple functional form of the Exponential distribution function, and the properties of $e$? Unique properties are scattered around all over Mathematics, and most of the time, they don't reflect some "deeper intuition" or "structure" - they just exist (thankfully).
Second, the square of a variable has very little relation with its level. Just consider $f(x) = x$ in, say, $[-2,\,2]$:
...or graph the standard normal density against the chi-square density: they reflect and represent totally different stochastic behaviors, even though they are so intimately related, since the second is the density of a variable that is the square of the first. The normal may be a very important pillar of the mathematical system we have developed to model stochastic behavior - but once you square it, it becomes something totally else.
Since $(X_1,X_2)$ is bi-variate normal, $X_1\sim N(\mu_1,\sigma_{11})$. For $Y=X_1^+$, the mean you derived is for special case that $\mu_1=0$, specifically, $$\operatorname{E}[Y]=\int_0^{\infty}xf_{X_1}(x)\,dx=\frac{\sigma_{11}}{\sqrt{2\pi}}.$$ Moreover, we can compute that $$\operatorname{E}[Y^2]=\int_0^{\infty}x^2f_{X_1}(x)\,dx=\frac{\sigma_{11}^2}{2}.$$ So $$\operatorname{Var}(Y)=\frac{\sigma_{11}^2}{2}\left(1-\frac{1}{\pi}\right).$$ This can also be derived from variance of half-normal distribution, note first $|X_1|=X_1^++X_1^-$, then $\operatorname{E}[X_1^+X_1^-]=0$ since whenever one r.v. is greater than $0$ the other is $0$, such that $$\operatorname{Cov}(X_1^+,X_1^-)=-\operatorname{E}[X_1^+]\operatorname{E}[X_1^-]=-\frac{\sigma_{11}^2}{2\pi};$$ finally, notice that $\operatorname{Var}(X_1^+)=\operatorname{Var}(X_1^-)$ by symmetry. It follows then $$\sigma_{11}^2\left(1-\frac{2}{\pi}\right)=\operatorname{Var}(|X_1|)=2\operatorname{Var}(X_1^+)-\frac{\sigma_{11}^2}{\pi},$$ which yields $$\operatorname{Var}(X_1^+)=\frac{\sigma_{11}^2}{2}\left(1-\frac{1}{\pi}\right)$$
For more general $\mu_1$, this entry about folded normal distribution may help.
Best Answer
Note that as $z^2=(-z)^2$ then $$\dfrac{\sigma^2}{\sqrt{2\pi}}\int_{-\infty}^\infty z^2e^{-z^2/2}dz=2\dfrac{\sigma^2}{\sqrt{2\pi}}\int_{0}^\infty z^2e^{-z^2/2}dz\,,$$ that is, as the mean is $z=0$, the function is symmetric around the $z=0$ axis, and the twice the $[0,\infty)$ area is the $(-\infty,\infty)$ area.
Now let $u=z^2/2$, then $du=zdz$ and $dz=\dfrac{du}{\sqrt{2u}}$ and
$$2\dfrac{\sigma^2}{\sqrt{2\pi}}\int_{0}^\infty z^2e^{-z^2/2}dz\rightarrow2\dfrac{\sigma^2}{\sqrt{\pi}}\int_{0}^\infty u^{1/2}e^{-u}du\,.$$
Finally, $\Gamma(z) = \int_0^\infty x^{z-1} e^{-x}\,dx$, thus our integral can be rewritten as $$2\dfrac{\sigma^2}{\sqrt{\pi}}\Gamma\left(\dfrac{3}{2}\right)=\sigma ^2\,\,.$$