Chapter 7: Q.7.47 (page 362)

Let X be a normal random variable with mean $μ$ and variance $σ^{2}$ . Use the results of Theoretical Exercise 7.46 to show that
$E [X^{n}] = \sum_{j = 0}^{[n / 2]} \frac{(\begin{matrix} n \\ 2 j \end{matrix}) μ^{n - 2 j} σ^{2 j} (2 j)!}{2 j j!}$
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 $E (X^{n}) = \sum_{j = 0}^{[n / 2]} (\begin{array}{l} n \\ 2 j \end{array}) σ^{2 j} \frac{(2 j)!}{2^{j} j!} μ^{n - 2 j}$

Step by step solution

Given information

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

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,

$E (X^{n}) = E ((μ + σ Z)^{n})$

$= \sum_{j = 0}^{n} (\begin{matrix} n \\ j \end{matrix}) σ^{j} E (Z^{j}) μ^{n - j}$

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

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

$= \sum_{j = 0}^{\infty} (\frac{{(t^{2} / 2)}^{j}}{j!})$

Solution

Now we need to compare and substitute the given,

$E (X^{n}) = \sum_{j \leq n, j even} (\begin{matrix} n \\ j \end{matrix}) σ^{j} E (Z^{j}) μ^{n - j}$

$= \sum_{j = 0}^{[n / 2} (\begin{array}{l} n \\ 2 j \end{array}) σ^{2 j} E (Z^{2 j}) μ^{n - 2 j}$

So as a result,

$E (X^{n}) = \sum_{j = 0}^{[n / 2} (\begin{array}{l} n \\ 2 j \end{array}) σ^{2 j} \frac{(2 j)!}{2^{j} j!} μ^{n - 2 j}$

Hence, $E (X^{n}) = \sum_{j = 0}^{[n / 2]} (\begin{array}{l} n \\ 2 j \end{array}) σ^{2 j} \frac{(2 j)!}{2^{j} j!} μ^{n - 2 j}$

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 $E (X^{n}) = \sum_{j = 0}^{[n / 2]} (\begin{array}{l} n \\ 2 j \end{array}) σ^{2 j} \frac{(2 j)!}{2^{j} j!} μ^{n - 2 j}$ .

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Let X be a normal random variable with mean μand variance σ2. Use the results of Theoretical Exercise 7.46 to show thatEXn=∑j=0[n/2] n2jμn−2jσ2j(2j)!2jj!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

Step by step solution

Given information

Solution

Solution

Final answer

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Most popular questions from this chapter

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Statistics

Probability and Statistics

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