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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[Zn+1]=nE[Zn-1].

(c) Find E[Z4].

Short Answer

Expert verified

Showing that ,

a)E[Zg(z)]=Zg(z)F(z)dz

=g(2)F(z)dz

b) E(zn+1)=zn+1F(2)dz

=nzn1F(z)dz

=nzn1F(n)dz+

c)E(z4)=3

Step by step solution

01

Given Information (part a)

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

02

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.

We need to showE(g(z))=E[zg(z)]

E[Zg(z)]=Zg(z)F(z)dz

=g(2)F(z)dz

E[zg(z)]=E[g(z)]

03

Final Answer (part a)

Showing that,

E[Zg(z)]=Zg(z)F(z)dz

=g(2)F(z)dz

04

Given Information (part b)

Show that E[Zn+1]=n E[Zn-1].

05

Explanation (part b)

Now,E(zn+1)=zn+1F(2)dz

=nzn1F(z)dz

=nzn1F(n)dz+

E[zn+1]=n[zn1]

06

Final Answer (part b)

shows that,

E(zn+1)=zn+1F(2)dz

=nzn1F(z)dz

=nzn1F(n)dz+

07

Given Information (part c)

Find E[Z4].

08

Explanation (part c)

Now findE(z4)

E(z4)=E(z31)

=3E(z2)

=3{vari(z)+(E(z)r)}

=3[1+0]

=3(1)

E(z4)=3

09

Final Answer (part c)

E(z4)=3

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