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$\mathrm{Ï•}=$standard normal distribution function.

$X=$normal random variable with mean $\mathrm{μ}$and variance 2

role="math" localid="1647418706310" $Z=$standard normal random variable that is independent of X

$I=\left\{\begin{array}{c}1,\text{if}\mathrm{Z}<\mathrm{X}\\ 0,\text{if}Zâ‰\yen \mathrm{X}\end{array}\right.$

We are going to show that $E[Iâˆ\pounds X=x]=\mathrm{Φ}\left(x\right)$

So,

$E[Iâˆ\pounds X=x]=(1×P\{Z<Xâˆ\pounds X=x\left\}\right)+(0×P\{Zâ‰\yen Xâˆ\pounds X=x\left\}\right)$

$=P\{Z<xâˆ\pounds X=x\}$

$=P\{Z<x\}$

Now take the given definition as $\mathrm{Φ}\left(x\right)$is standard normal distribution function, one has that:

So that,

$P\{Z<x\}=\mathrm{Φ}\left(x\right)$

$E[Iâˆ\pounds X=x]=\mathrm{Φ}\left(x\right)$

It has been now shown that the$E[Iâˆ\pounds X=x]=\mathrm{Φ}\left(x\right)$.

$\mathrm{Ï•}=$standard normal distribution function.

$X=$normal random variable with mean $\mathrm{μ}$and variance 2

$Z=$standard normal random variable that is independent of X

$I=\left\{\begin{array}{c}1,\text{if}\mathrm{Z}<\mathrm{X}\\ 0,\text{if}\mathrm{Z}â‰\yen \mathrm{X}\end{array}\right.$

Here we need to show that $E\left[\mathrm{Φ}\right(x\left)\right]=P\{Z<X\}$

From the above result,

$E[Iâˆ\pounds X]=\mathrm{Φ}\left(X\right)$

Moreover, one has the following relations:

$E\left[I\right]=E\left[E\right[Iâˆ\pounds X\left]\right]$

$=E\left[\mathrm{Φ}\right(X\left)\right]−−−−−\left(1\right)$

So,

$E\left[I\right]=\left(P\right\{I=1\}×1)+(0×P\{I=0\left\}\right)$

$=P\{Z<X\}×1+P\{Zâ‰\yen X\}×0$

$=P\{Z<X\}−−−−−−\left(2\right)$

Now we need to combine The equation (1) and (2) That will get.

$E\left[\mathrm{Φ}\right(X\left)\right]=P\{Z<X\}$

Now it has been shown that$E\left[\mathrm{Φ}\right(X\left)\right]=P\{Z<X\}$.

$\mathrm{Ï•}=$standard normal distribution function.

$X=$normal random variable with mean $\mathrm{μ}$and variance 2

$Z=$standard normal random variable that is independent of X

$I=\left\{\begin{array}{c}1,\text{if}Z<\mathrm{X}\\ 0,\text{if}Zâ‰\yen \mathrm{X}\end{array}\right.$

We need to show that $E\left[\mathrm{Φ}\right(x\left)\right]=\mathrm{Φ}\left(\frac{\mathrm{μ}}{\sqrt{2}}\right)$

$P\{X>Z\}=P\{X−Z>0\}$

$=P\left\{\frac{X−Z−\mathrm{μ}}{\sqrt{2}}>\frac{−\mathrm{μ}}{\sqrt{2}}\right\}$

$=1−\mathrm{Φ}\left(\frac{−\mathrm{μ}}{\sqrt{2}}\right)$

$=\mathrm{Φ}\left(\frac{\mathrm{μ}}{\sqrt{2}}\right)$

Therefore the result is,

$E\left[\mathrm{Φ}\right(X\left)\right]=\mathrm{Φ}\left(\frac{\mathrm{μ}}{\sqrt{2}}\right)$

Now it has been shown that$E\left[\mathrm{Φ}\right(X\left)\right]=\mathrm{Φ}\left(\frac{\mathrm{μ}}{\sqrt{2}}\right)$.