Starting with etX=1+tX+t2X22!+t3X33!+…+tnXnn!+…show that (a) M(t)=EetX=1+tE[X]+t2EX22!+…+tnEXnn!+…(b) Use (a) to show that dndtnM(t)t=0=EXn

The answer isThe expectation value of etX is localid="1647527908191" EetX=1+t⋅E[X]+t22!EX2+t33!EX3+…+tnn!EXn+….It has been shown that dndtnM(t)t=0=EXn.

Q.7.32 - A First Course in Probability

Q.7.32

Question

Starting with

$e^{t X} = 1 + t X + \frac{t^{2} X^{2}}{2!} + \frac{t^{3} X^{3}}{3!} + \dots + \frac{t^{n} X^{n}}{n!} + \dots$

show that

(a) $M (t) = E [e^{t X}] = 1 + t E [X] + \frac{t^{2} E [X^{2}]}{2!} + \dots + \frac{t^{n} E [X^{n}]}{n!} + \dots$

(b) Use (a) to show that ${\frac{d^{n}}{d t^{n}} M (t)|}_{t = 0} = E [X^{n}]$

Step-by-Step Solution

Verified

Answer

The answer is

The expectation value of $e^{t X}$ is $E [e^{t X}] = 1 + t \cdot E [X] + \frac{t^{2}}{2!} E [X^{2}] + \frac{t^{3}}{3!} E [X^{3}] + \dots + \frac{t^{n}}{n!} E [X^{n}] + \dots$ .
It has been shown that ${\frac{d^{n}}{d t^{n}} M (t)|}_{t = 0} = E [X^{n}]$ .

1Step 1: Given information (Part a)

First equation $e^{t X} = 1 + t X + \frac{t^{2} X^{2}}{2!} + \frac{t^{3} X^{3}}{3!} + \dots + \frac{t^{n} X^{n}}{n!} + \dots$

Second equation $M (t) = E [e^{t X}] = 1 + t E [X] + \frac{t^{2} E [X^{2}]}{2!} + \dots + \frac{t^{n} E [X^{n}]}{n!} + \dots$

2Step 2: Solution (Part a)

We need to find that the expectation value of $e^{t x}$

So,

$M (t) = \int (e^{t x}) f (x) d x$

$M (t) = E [e^{ı X}]$

$\Rightarrow E [e^{u X}] = E [1 + t X + \frac{t^{2} X^{2}}{2!} + \frac{t^{3} X^{3}}{3!} + \dots + \frac{t^{n} X^{n}}{n!} + \dots]$

$\Rightarrow E [e^{t X}] = E [1] + E [t X] + E [\frac{t^{2} X^{2}}{2!}] + E [\frac{t^{3} X^{3}}{3!}] + \dots + E [\frac{t^{n} X^{n}}{n!}] + \dots$

$E [e^{ι X}] = 1 + t \cdot E [X] + \frac{t^{2}}{2!} E [X^{2}] + \frac{t^{3}}{3!} E [X^{3}] + \dots + \frac{t^{n}}{n!} E [X^{n}] + \dots$

3Step 3: Final answer (Part a)

The expectation value of $e^{t X}$ is $E [e^{' X}] = 1 + t \cdot E [X] + \frac{t^{2}}{2!} E [X^{2}] + \frac{t^{3}}{3!} E [X^{3}] + \dots + \frac{t^{n}}{n!} E [X^{n}] + \dots$

4Step 4: Given information (Part b)

First equation $e^{t X} = 1 + t X + \frac{t^{2} X^{2}}{2!} + \frac{t^{3} X^{3}}{3!} + \dots + \frac{t^{n} X^{n}}{n!} + \dots$

Second equation ${\frac{d^{n}}{d t^{n}} M (t)|}_{t = 0} = E [X^{n}]$

5Step 5: Solution (Part b)

We need to show that ${\frac{d^{n}}{d t^{n}} M (t)|}_{t = 0} = E [X^{n}]$

${\frac{d^{n}}{d t^{n}} M (t)|}_{t = 0} = \frac{d^{n}}{d t^{n}} {(1 + t \cdot E [X] + \frac{t^{2}}{2!} E [X^{2}] + \frac{t^{3}}{3!} E [X^{3}] + \dots + \frac{t^{n}}{n!} E [X^{n}] + \dots)}_{t = 0}$

${\frac{d^{n}}{d t^{n}} M (t)|}_{t = 0} = \frac{d^{n}}{d t^{n}} (1)_{t = 0} + \frac{d^{n}}{d t^{n}} (t \cdot E [X])_{t = 0} + \frac{d^{n}}{d t^{n}} {(\frac{t^{2}}{2!} E [X^{2}])}_{t = 0} + \frac{d^{n}}{d t^{n}} {(\frac{t^{3}}{3!} E [X^{3}])}_{t = 0} + \dots$

$+ \frac{d^{n}}{d t^{n}} {(\frac{t^{n}}{n!} E [X^{n}])}_{t = 0} + \dots$

6Step 6: Solution (Part b)

The terms with $t^{i}$ for $i < n$ vanish because the $n^{th}$ derivative is 0 . On the other hand, terms with $t^{i}$ for $i > n$ vanish because the derivative is evaluated for $t = 0$ . Hence, only the $n^{t h}$ term remains.

${\frac{d^{n}}{d t^{n}} M (t)|}_{t = 0} = \frac{n!}{n!} E [X^{n}]$

${\frac{d^{n}}{d t^{n}} M (t)|}_{t = 0} = E [X^{n}]$

7Step 7: Final answer (Part b)

It has been shown that in the solution that is ${\frac{d^{n}}{d t^{n}} M (t)|}_{t = 0} = E [X^{n}]$

Q.7.31

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Q.7.31

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Q.7.29

Suppose that X and Y are both Bernoulli random variables. Show that X and Y are independent if and only if Cov(X, Y) = 0.

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