Question-36
Let \(Y=Z^{2}\), \(Z=\frac{X-\mu }{\sigma }\) , where \(X\) is a normal random variable with mean \(\mu\) and variance \(\sigma ^{2}\). The variance of \(Y\) is
We will use the result \(\sigma _{X}^{2} =\mathbb{E}\left[ X^{2}\right] -\mathbb{E}[ X]^{2}\) repeatedly. First, we note that \(Z\sim \mathcal{N}( 0,1)\), the standard normal random variable. Next:
\[ \begin{aligned} \sigma _{Y}^{2} & =\mathbb{E}\left[ Y^{2}\right] -\mathbb{E}[ Y]^{2}\\ & =\mathbb{E}\left[ Z^{4}\right] -\mathbb{E}\left[ Z^{2}\right]^{2} \end{aligned} \]
We see that \(\sigma _{Z}^{2} =\mathbb{E}\left[ Z^{2}\right] -\mathbb{E}[ Z]^{2}\). Since \(\mathbb{E}[ Z] =0\) and \(\sigma _{Z} =1\), we have \(\mathbb{E}\left[ Z^{2}\right] =1\). This leaves us with \(\mathbb{E}\left[ Z^{4}\right]\), the fourth moment of \(Z\), which is \(3\). Therefore, the variance of \(Y\) is \(2\).
One can also look at \(Z^{2}\) as a Chi-squared random variable with one degree of freedom. The variance of a Chi-squared random variable with \(k\) degrees of freedom is \(2k\). Here, \(k=1\), so the variance is \(2\). Recall that a Chi-squared random variable with \(k\) degrees of freedom is defined as the sum of squares of \(k\) independent standard normal random variables.
Read on if you want to know how the fourth moment is computed from first principles.
We can go back to first principles here:
\[ \begin{aligned} \mathbb{E}\left[ Z^{4}\right] & =\frac{1}{\sqrt{2\pi }} \ \int\limits _{-\infty }^{\infty } z^{4} \cdot e^{-\frac{z^{2}}{2}} dz \end{aligned} \]
Now, recall that from the product rule for derivatives we can get hold of the formula for integration by parts:
\[ \begin{aligned} \int\limits _{a}^{b} d( uv) & =\int\limits _{a}^{b} udv+\int\limits _{a}^{b} vdu\\ \Longrightarrow \int\limits _{a}^{b} udv & =\int\limits _{a}^{b} d( uv) -\int\limits _{a}^{b} vdu\\ & =[ uv]_{a}^{b} -\int\limits _{a}^{b} vdu \end{aligned} \] Let us set \(u=z^{3}\) and \(dv=z \cdot e^{\frac{-z^{2}}{2}}\):
\[ \begin{aligned} \frac{1}{\sqrt{2\pi }} \ \int\limits _{-\infty }^{\infty } z^{4} \cdot e^{-\frac{z^{2}}{2}} dz & =\frac{1}{\sqrt{2\pi }} \ \left[ z^{3} \cdot e^{\frac{-z^{2}}{2}} \cdot ( -1)\right]_{-\infty }^{\infty }\\ & -\frac{1}{\sqrt{2\pi }} \ \int\limits _{-\infty }^{\infty } 3z^{2} \cdot e^{\frac{-z^{2}}{2}} \cdot ( -1) dz\\ & =3 \end{aligned} \]
Here, the first term is zero since \(\lim\limits _{z\rightarrow \pm \infty } z^{3} \cdot e^{\frac{-z^{2}}{2}} =0\). The second term happens to be the second moment of \(Z\) (which is \(1\)) scaled by \(3\).