Use the identity of Theoretical Exercise 5.5 to derive E[X2] when X is an exponential random variable with parameter λ.

Short Answer

Expert verified

Thus,

02xP(X>x)dx=02xeλxdx=20xeλxdx

Use partial integration . Letu=xanddv=eλx. Hence, we have thatdu=dxandv=eλxλ. Thus

20xeλxdx=2xeλxλ|0+2λ0eλxdx=2λ(eλxλ)|0=2λ2

Step by step solution

01

Given Information 

Derive E[X2] when X is an exponential random variable with parameter λ.

02

Explanation

From exercise 5, we have that.

E(X2)=02xP(X>x)dx

Now, we are given thatX~Expo(λ). Hence, we have that

P(X>x)=1P(Xx)=1(1eλx)=eλx

Thus,

02xP(X>x)dx=02xeλxdx=20xeλxdx

Use partial integration . Letu=xanddv=eλx. Hence, we have thatdu=dxandv=eλxλ. Thus

20xeλxdx=2xeλxλ|0+2λ0eλxdx=2λ(eλxλ)|0=2λ2

Hence, we have proved that

E(X2)=2λ2

03

Final Answer 

Thus,

02xP(X>x)dx=02xeλxdx=20xeλxdx

Use partial integration . Letu=xanddv=eλx. Hence, we have thatdu=dxandv=eλxλ. Thus

20xeλxdx=2xeλxλ|0+2λ0eλxdx=2λ(eλxλ)|0=2λ2

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