91影视

The required integrals are given by:

The given Gaussian probability distribution is assumed to be valid over the whole line

i.e.

${\mathrm{蚁}}_{\left(\mathrm{x}\right)}={\mathrm{Ae}}^{鈭�\mathrm{位}{(\mathrm{x}鈭�\mathrm{a})}^2}$ $鈭�\mathrm{鈭�}鈮�\mathrm{x}鈮�\mathrm{鈭�}$

Calculating for A by normalizing the function,

$\begin{array}{c}\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{蚁}}_{\left(\mathrm{x}\right)}\mathrm{dx}=1\\ \underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{Ae}}^{鈭�\mathrm{位}{(\mathrm{x}鈭�\mathrm{a})}^2}\mathrm{dx}=1\end{array}$

Assuming, $\mathrm{y}=\mathrm{x}鈭�\mathrm{a}$

Hence, the equation becomes

$\mathrm{A}\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{e}}^{鈭�{\mathrm{位测}}^2}\mathrm{dy}=1$

$\mathrm{A}\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{e}}^{\frac{鈭�{\mathrm{y}}^2}{{(1/\sqrt{\mathrm{位}})}^2}}\mathrm{dy}=1$

This integral can be taken from $0\mathrm{to}\mathrm{鈭�}$, multiplying with a factor 2, since it鈥檚 an even function of y.

$2\mathrm{A}\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{e}}^{\frac{鈭�{\mathrm{y}}^2}{{(1/\sqrt{\mathrm{位}})}^2}}\mathrm{dy}=1$

Now, from equation 1, putting $\mathrm{n}=0$

Thus, the value of A is $\sqrt{\frac{\mathrm{位}}{\mathrm{蟺}}}$.

Hence, the Normalised probability distribution comes out to be

${\mathrm{蚁}}_{\left(\mathrm{x}\right)}=\sqrt{\frac{\mathrm{位}}{\mathrm{蟺}}}{\mathrm{e}}^{鈭�\mathrm{位}{(\mathrm{x}鈭�\mathrm{a})}^2}$ $鈭�\mathrm{鈭�}鈮�\mathrm{x}鈮�\mathrm{鈭�}$

First, we solve for $鉄�{\mathrm{x}}^2鉄�$

role="math" localid="1655379313238"

Since, the integral of an odd function over a symmetric interval is zero

i.e. $\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{ye}}^{鈭�{\mathrm{位测}}^2}\mathrm{dy}=0$

Therefore,

Thus, value of $鉄�x鉄�$ is $\mathrm{a}$.

Now, for $鉄�{\mathrm{x}}^2鉄�$

Thus, value of $鉄�{\mathrm{x}}^2鉄�$ is $\frac{1}{2\mathrm{位}}+{\mathrm{a}}^2$.

The standard deviation can be calculated as,

$\begin{array}{l}\mathrm{蟽}=\sqrt{鉄�{\mathrm{x}}^2鉄�鈭�{鉄�\mathrm{x}鉄�}^2}\\ \mathrm{蟽}=\sqrt{(\frac{1}{2\mathrm{位}}+{\mathrm{a}}^2)鈭�{\mathrm{a}}^2}\\ \mathrm{蟽}=\sqrt{\frac{1}{2\mathrm{位}}}\end{array}$

Thus, value of 蟽 is $\sqrt{\frac{1}{2\mathrm{位}}}$.

Figure (1)

Here,