Question

Let Z be a standard normal random variable Z, and let g be a differentiable function with derivative g'.

(a) Show that E[g'(Z)]=E[Zg(Z)];

(b) Show that E[Z^n+$1$]=nE[Z^n-$1$].

(c) Find E[Z^$4$].

Short answer

Showing that ,

a)$E\left[Zg\right(z\left)\right]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}Zg\left(z\right)F\left(z\right)dz$

$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{g}^{\mathrm{\prime }}\left(2\right)F\left(z\right)dz$

b) $E\left(z^{n+1}\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{z}^{n+1}F\left(2\right)dz$

$={\int }_{\mathrm{\infty }}^{\mathrm{\infty }}n{z}^{n-1}F\left(z\right)dz$

$=n{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{z}^{n-1}F\left(n\right)dz+$

c)$E\left(z^4\right)=3$

Step-by-step solution

Given Information (part a)

Show that E[g'(Z)]=E[Zg(Z)]

Step 2: Explanation (part a)

Let us Consider ' Z " to be a standard normal random variable z a let 'g' be the differentiable function with derivative g.

$\text{We need to show}E\left(g\right(z\left)\right)=E\left[z_g\right(z\left)\right]$

$E\left[Zg\right(z\left)\right]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}Zg\left(z\right)F\left(z\right)dz$

$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{g}^{\mathrm{\prime }}\left(2\right)F\left(z\right)dz$

$E\left[zg\right(z\left)\right]=E\left[g\right(z\left)\right]$

Final Answer (part a)

Showing that,

$E\left[Zg\right(z\left)\right]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}Zg\left(z\right)F\left(z\right)dz$

$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{g}^{\mathrm{\prime }}\left(2\right)F\left(z\right)dz$

Given Information (part b)

Show that E[Z^n+$1$]=n E[Z^n-$1$].

Explanation (part b)

$\text{Now,}E\left(z^{n+1}\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{z}^{n+1}F\left(2\right)dz$

$={\int }_{\mathrm{\infty }}^{\mathrm{\infty }}n{z}^{n-1}F\left(z\right)dz$

$=n{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{z}^{n-1}F\left(n\right)dz+$

$E\left[z^{n+1}\right]=n\in \left[z^{n-1}\right]$

Final Answer (part b)

shows that,

$E\left(z^{n+1}\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{z}^{n+1}F\left(2\right)dz$

$={\int }_{\mathrm{\infty }}^{\mathrm{\infty }}n{z}^{n-1}F\left(z\right)dz$

$=n{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{z}^{n-1}F\left(n\right)dz+$

Given Information (part c)

Find E[Z^$4$].

Explanation (part c)

$\text{Now find}E\left(z^4\right)$

$E\left(z^4\right)=E\left(z^{3-1}\right)$

$=3E\left(z^2\right)$

$=3\{\mathrm{vari}(z)+(E\left(z\right)r\left)\right\}$

$=3[1+0]$

$=3\left(1\right)$

$E\left(z^4\right)=3$

Final Answer (part c)

$E\left(z^4\right)=3$