Q.7.47 Let X be a normal random variabl... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 7: Q.7.47 (page 362)

Let X be a normal random variable with mean $渭$ and variance $蟽^{2}$ . Use the results of Theoretical Exercise 7.46 to show that
$E [X^{n}] = {鈭�}_{j = 0}^{[n / 2]} 鈥� \frac{(\begin{matrix} n \\ 2 j \end{matrix}) 渭^{n 鈭� 2 j} 蟽^{2 j} (2 j)!}{2 j j!}$
In the preceding equation, $[n / 2]$ is the largest integer less than or equal to $n / 2$ . Check your answer by letting $n = 1$ and $n = 2$ .

Short Answer

Expert verified

It has been shown that $E (X^{n}) = {鈭�}_{j = 0}^{[n / 2]} 鈥� (\begin{array}{l} n \\ 2 j \end{array}) 蟽^{2 j} \frac{(2 j)!}{2^{j} j!} 渭^{n 鈭� 2 j}$

Step by step solution

Given information

X be a normal random variable with mean $渭$ and variance $蟽^{2}$ .

In the preceding equation, $[n / 2]$ is the largest integer less than or equal to $n / 2$ .

Solution

We need to use the binomial theorem to obtain the result,

$E (X^{n}) = E ((渭 + 蟽 Z)^{n})$

$= {鈭�}_{j = 0}^{n} 鈥� (\begin{matrix} n \\ j \end{matrix}) 蟽^{j} E (Z^{j}) 渭^{n 鈭� j}$

From theoretical exercise 46 , we need to obtain the following result:

$E [e^{t z}] = e^{\frac{t^{2}}{2}}$

$= {鈭�}_{j = 0}^{鈭�} 鈥� (\frac{{(t^{2} / 2)}^{j}}{j!})$

Solution

Now we need to compare and substitute the given,

$E (X^{n}) = \underset{j 鈮� n, j even}{鈭�} 鈥� (\begin{matrix} n \\ j \end{matrix}) 蟽^{j} E (Z^{j}) 渭^{n 鈭� j}$

$= {鈭�}_{j = 0}^{[n / 2} 鈥� (\begin{array}{l} n \\ 2 j \end{array}) 蟽^{2 j} E (Z^{2 j}) 渭^{n 鈭� 2 j}$

So as a result,

$E (X^{n}) = {鈭�}_{j = 0}^{[n / 2} 鈥� (\begin{array}{l} n \\ 2 j \end{array}) 蟽^{2 j} \frac{(2 j)!}{2^{j} j!} 渭^{n 鈭� 2 j}$

Hence, $E (X^{n}) = {鈭�}_{j = 0}^{[n / 2]} 鈥� (\begin{array}{l} n \\ 2 j \end{array}) 蟽^{2 j} \frac{(2 j)!}{2^{j} j!} 渭^{n 鈭� 2 j}$

Where, $[n / 2]$ is the greatest integer less than or equal to $n / 2$ .

Final answer

From the above calculation it has been shown that $E (X^{n}) = {鈭�}_{j = 0}^{[n / 2]} 鈥� (\begin{array}{l} n \\ 2 j \end{array}) 蟽^{2 j} \frac{(2 j)!}{2^{j} j!} 渭^{n 鈭� 2 j}$ .

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