Answer to Question #208888 in Statistics and Probability for MD. M. H. Akash

Question

Find the mean and variance of normal distribution using density function

Answer 1

Let $X$ is distributed according to the probability density function

$$f(x) = \frac{1}{{\sqrt {2\pi } \sigma }}{e^{ - \frac{{{{(x - \mu )}^2}}}{{2{\sigma ^2}}}}}$$

Then it's moment generation function

$${M_X}(\lambda ) = \mathbb{E}({e^{\lambda X}}) = \int\limits_{ - \infty }^{ + \infty } {f(x){e^{\lambda x}}dx}$$

Let's evaluate this integral

$$\int\limits_{ - \infty }^{ + \infty } {f(x){e^{\lambda x}}dx} = \frac{1}{{\sqrt {2\pi } \sigma }}\int\limits_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{(x - \mu )}^2}}}{{2{\sigma ^2}}} + \lambda x}}dx}$$

At the first, complete the square in the power of the exponent

$$- \frac{{{{(x - \mu )}^2}}}{{2{\sigma ^2}}} + tx = - \frac{1}{{2{\sigma ^2}}}{(x - (\mu + \lambda {\sigma ^2}))^2} + \mu \lambda + \frac{{{\lambda ^2}{\sigma ^2}}}{2}$$

Then change the variable $x - (\mu + \lambda {\sigma ^2}) = \xi$ , then $dx = d\xi$ , limits do not change

$${M_X}(\lambda ) = \frac{1}{{\sqrt {2\pi } \sigma }}{e^{\mu \lambda + \frac{{{\lambda ^2}{\sigma ^2}}}{2}}}\int\limits_{ - \infty }^{ + \infty } {{e^{ - \frac{1}{{2{\sigma ^2}}}{\xi ^2}}}d\xi }$$

We need only to evaluate $I = \int\limits_{ - \infty }^{ + \infty } {{e^{ - \frac{1}{{2{\sigma ^2}}}{\xi ^2}}}d\xi }$ . Let rewrite it's square as a double integral at the whole ${\mathbb{R}^2}$ plane

$${I^2} = \int\limits_{ - \infty }^{ + \infty } {{e^{ - \frac{1}{{2{\sigma ^2}}}{\xi ^2}}}d\xi } \cdot \int\limits_{ - \infty }^{ + \infty } {{e^{ - \frac{1}{{2{\sigma ^2}}}{\eta ^2}}}d\eta } = \iint\limits_{{R^2}} {{e^{ - \frac{1}{{2{\sigma ^2}}}({\xi ^2} + {\eta ^2})}}d\xi d\eta }$$

And choose polar coordinates $\xi = r\cos \varphi$ , $\eta = r\sin \varphi$ and $d\xi d\eta = rdrd\varphi$ .

Then

$${I^2} = \iint\limits_{{\mathbb{R}^2}} {{e^{ - \frac{{{r^2}}}{{2{\sigma ^2}}}}}rdrd\varphi } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^{ + \infty } {r{e^{ - \frac{{{r^2}}}{{2{\sigma ^2}}}}}dr}$$

and

$${I^2} = 2\pi \int\limits_0^{ + \infty } {r{e^{ - \frac{{{r^2}}}{{2{\sigma ^2}}}}}dr} = 2\pi {\sigma ^2}\int\limits_0^{ + \infty } {{e^{ - \frac{{{r^2}}}{{2{\sigma ^2}}}}}d} (\frac{{{r^2}}}{{2{\sigma ^2}}}) = - 2\pi {\sigma ^2}\left. {{e^{ - \frac{{{r^2}}}{{2{\sigma ^2}}}}}} \right|_0^{ + \infty } = 2\pi {\sigma ^2}$$

We get

$$\int\limits_{ - \infty }^{ + \infty } {{e^{ - \frac{1}{{2{\sigma ^2}}}{\xi ^2}}}d\xi } = \sqrt {2\pi } \sigma$$

and thus

$${M_X}(\lambda ) = {e^{\mu \lambda + \frac{{{\lambda ^2}{\sigma ^2}}}{2}}}$$

Now, using that

To calculate expected value and variance we can use that

$$\mathbb{E}({X^n}) = M_X^{(n)}(0)$$

Let's calculate the expected value (mean value)

$$\mathbb{E}X = {\left. {\frac{{dM_X}}{{d\lambda }}} \right|_{\lambda = 0}} = {\left. {(\mu + \lambda {\sigma ^2}){e^{\mu \lambda + \frac{{{\lambda ^2}{\sigma ^2}}}{2}}}} \right|_{\lambda = 0}} = \mu$$

To calculate variance, let's calculate the second moment at first

$$\mathbb{E}{X^2} = {\left. {\frac{{{d^2}M_X}}{{d{\lambda ^2}}}} \right|_{\lambda = 0}} = [{\left. {{\sigma ^2}{e^{\mu \lambda + \frac{{{\lambda ^2}{\sigma ^2}}}{2}}} + {{(\mu + \lambda {\sigma ^2})}^2}{e^{\mu \lambda + \frac{{{\lambda ^2}{\sigma ^2}}}{2}}}]} \right|_{\lambda = 0}} = {\mu ^2} + {\sigma ^2}$$

Now calculate the variance

$$\operatorname{Var} X = E{X^2} - {E^2}X = {\sigma ^2}$$