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Chapter 5: Q.5.4 (page 215)

Prove Corollary2.1.

Short Answer

Expert verified

Therefore,

E(ax+b)=aE(x)+b

Hence Proved.

Step by step solution

01

Given Information.

E(ax+b)=aE(x)+b

02

Explanation.

E(ax+b)=aE(x)+b

E(ax+b)=鈭�0鈭�(ax+b)f(x)dx

03

Explanation.

=鈭�0鈭�ax.f(x)dx+鈭�0鈭�b.f(x)dx

=a鈭�0鈭�x.f(x)dx+b.鈭�0鈭�f(x)dx

=aE(x)+b

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Most popular questions from this chapter

5.6. ComputeE[X]if Xhas a density function given by

(a)f(x)=14xe-x2x>00otherwise;

(b)f(x)=c(1鈭�x2)-1<x<10otherwise;

(c)f(x)=5x2x>50x鈮�5.

Use the result that for a nonnegative random variableY,

EY=鈭�0鈭�PY>tdt

to show that for a nonnegative random variableX,

EXn=鈭�0鈭�nxn-1PX>xdx

Hint: Start with

EXn=鈭�0鈭�PXn>tdt

and make the change of variablest=xn.

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(a)What is the probability that the machine will still be working 6years after being purchased?

(b) If it is still working 6years after being purchased, what is the conditional probability that it will fail within the next

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