Question

What is the variance of the standard normal distribution?

Answer 1

Hint: To find the variance of the standard normal distribution, we will use the formula $\text{Var}\left[ X \right]=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}$ . We can find $E\left[ {{X}^{2}} \right]$ using the formula $E\left[ {{X}^{2}} \right]=\int\limits_{-\infty }^{\infty }{{{x}^{2}}{{f}_{x}}\left( x \right)dx}$ and substituting for ${{f}_{x}}\left( x \right)=\dfrac{1}{\sqrt{2\pi }}{{e}^{-\dfrac{1}{2}{{x}^{2}}}}$ . Then, we have to integrate by substitution method and apply the properties of Gamma functions. We can find $E\left[ X \right]$ using the definition of standard normal distribution which will result in a value 0. Then, we have to substitute the values in the variance equation and solve.

Complete step by step solution:
Let us first see what a standard normal distribution is. The standard normal distribution is one of the forms of the normal distribution. It occurs when a normal random variable has a mean equal to zero and a standard deviation equal to one.
Let us find the variance of standard normal random variable X. We know that variance of X is given by
$\text{Var}\left[ X \right]=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}...\left( A \right)$
We know that $E\left[ {{X}^{2}} \right]=\int\limits_{-\infty }^{\infty }{{{x}^{2}}{{f}_{x}}\left( x \right)dx}$
Where ${{f}_{x}}\left( x \right)=\dfrac{1}{\sqrt{2\pi }}{{e}^{-\dfrac{1}{2}{{x}^{2}}}}$ is the Probability Density Function of a standard normal distribution. Let us substitute this value in the above equation.
$\Rightarrow E\left[ {{X}^{2}} \right]=\dfrac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{{{x}^{2}}{{e}^{-\dfrac{1}{2}{{x}^{2}}}}dx}$
We can rewrite the above integral as
$E\left[ {{X}^{2}} \right]=\dfrac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{x{{e}^{-\dfrac{1}{2}{{x}^{2}}}}xdx}...\left( i \right)$
Let us assume $y=\dfrac{{{x}^{2}}}{2}...\left( ii \right)$ . We have to differentiate equation (ii) with respect to x.
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\times 2x \\
 & \Rightarrow dy=xdx...\left( iii \right) \\
\end{align}$
Now, from equation (ii), we can find the value of x as
$\Rightarrow x=\sqrt{2y}...\left( iv \right)$
Let us substitute (ii), (iii) and (iv) in equation (i).
$\Rightarrow E\left[ {{X}^{2}} \right]=\dfrac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{\sqrt{2y}{{e}^{-y}}dy}$
Let us take the constants outside.
\[\begin{align}
  & \Rightarrow E\left[ {{X}^{2}} \right]=\dfrac{\sqrt{2}}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{{{y}^{\dfrac{1}{2}}}{{e}^{-y}}dy} \\
 & \Rightarrow E\left[ {{X}^{2}} \right]=\dfrac{1}{\sqrt{\pi }}\int\limits_{-\infty }^{\infty }{{{y}^{\dfrac{1}{2}}}{{e}^{-y}}dy} \\
\end{align}\]
We know that $\int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=2\int\limits_{0}^{\infty }{f\left( x \right)dx}$ . Therefore, we can write the above integral as
\[\Rightarrow E\left[ {{X}^{2}} \right]=\dfrac{2}{\sqrt{\pi }}\int\limits_{0}^{\infty }{{{y}^{\dfrac{1}{2}}}{{e}^{-y}}dy}...\left( v \right)\]
We can see that \[\int\limits_{0}^{\infty }{{{y}^{\dfrac{1}{2}}}{{e}^{-y}}dy}\] is a Gamma function of the form $\Gamma \left( a \right)=\int\limits_{0}^{\infty }{{{y}^{a-1}}{{e}^{-y}}dy}$ for $a>0$ . We can write equation (v) as
\[\Rightarrow E\left[ {{X}^{2}} \right]=\dfrac{2}{\sqrt{\pi }}\int\limits_{0}^{\infty }{{{y}^{\left( \dfrac{1}{2}+1 \right)-1}}{{e}^{-y}}dy}\]
Now, let us convert the integral into Gamma function.
\[\Rightarrow E\left[ {{X}^{2}} \right]=\dfrac{2}{\sqrt{\pi }}\Gamma \left( \dfrac{1}{2}+1 \right)\]
We know that $\Gamma \left( z+1 \right)=z\Gamma \left( z \right)$ . Therefore, we can write the above equation as
\[\Rightarrow E\left[ {{X}^{2}} \right]=\dfrac{2}{\sqrt{\pi }}\dfrac{1}{2}\Gamma \left( \dfrac{1}{2} \right)\]
We know that \[\Gamma \left( \dfrac{1}{2} \right)=\sqrt{\pi }\] . Therefore, the above equation becomes
\[\Rightarrow E\left[ {{X}^{2}} \right]=\dfrac{2}{\sqrt{\pi }}\dfrac{1}{2}\times \sqrt{\pi }\]
Let us cancel the common terms.
\[\Rightarrow E\left[ {{X}^{2}} \right]=1\]
Let us substitute the above value in equation (A).
$\text{Var}\left[ X \right]=1-E{{\left[ X \right]}^{2}}$
We know that for a standard normal distribution, mean or expectation is 0. Therefore, the above equation becomes
$\begin{align}
  & \Rightarrow \text{Var}\left[ X \right]=1-{{0}^{2}} \\
 & \Rightarrow \text{Var}\left[ X \right]=1 \\
\end{align}$
Therefore, the variance of the standard normal distribution is 1.

Note: Students must know the mean of standard normal distribution to find the variance. They must also know to integrate functions and also the values of the integral of basic functions. They must also know about the Gamma distribution or function and few properties related to these.