Solution Found!
Let Z be a standard normal random variable.
Chapter 4, Problem 199SE(choose chapter or problem)
Let Z be a standard normal random variable.
a Show that the expected values of all odd integer powers of 𝑍 are 0. That is, if \(i=1\), 2, . . . ,
show that \(E\left(Z^{2 i-1}\right)=0\). [Hint: A function \(\mathrm{g}(\cdot)\) is an odd function if, for all \(y,\ g(-y)=-g(y)\). For any odd function \(g(y),\ \int_{-\infty}^{\infty}g(y)\ dy=0\), if the integral exists.]
b If \(i=1\), 2, . . . , show that
\(E\left(Z^{2 i}\right)=\frac{2^{i} \Gamma\left(i+\frac{1}{2}\right)}{\sqrt{\pi}}.\)
[Hint: A function \(h(\cdot)\) is an even function if, for all \(y,\ h(-y)=h(y)\). For any even function \(h(y),\ \int_{-\infty}^{\infty}h(y)\ dy=2\int_0^{\infty}h(y)\ dy\) if the integrals exist. Use this fact, make the change of variable \(w=z^{2} / 2\), and use what you know about the gamma function.]
c Use the results in part (b) and in Exercises 4.81(b) and 4.194 to derive \(E\left(Z^2\right),\ E\left(Z^4\right),\ E\left(Z^6\right),\text{ and }E\left(Z^8\right)\).
d If \(i=1\), 2, . . . , show that
\(E\left(Z^{2 i}\right)=\prod_{j=1}^{i}(2 j-1).\)
This implies that the ith even moment is the product of the first 𝑖 odd integers.
Equation Transcription:
Text Transcription:
i=1
E(Z^2i-1)=0
g(cdot)
y,g(-y)=-g(y)
g(y), integral -infinity to infinity g(y) dy=0
i=1
E(Z^2i)=2^i Gamma(i+1over 2)over sqrt pi.
h(cdot)
y,h(-y)=h(y)
h(y), integral -infinity to infinity h(y) dy=2 integral 0 to infinity h(y) dy
w=z^2/2
E(Z^2), E(Z^4), E(Z^6), and E(Z^8)
i=1
E(Z^2i)=product j=1 to i (2j-1).
Questions & Answers
QUESTION:
Let Z be a standard normal random variable.
a Show that the expected values of all odd integer powers of 𝑍 are 0. That is, if \(i=1\), 2, . . . ,
show that \(E\left(Z^{2 i-1}\right)=0\). [Hint: A function \(\mathrm{g}(\cdot)\) is an odd function if, for all \(y,\ g(-y)=-g(y)\). For any odd function \(g(y),\ \int_{-\infty}^{\infty}g(y)\ dy=0\), if the integral exists.]
b If \(i=1\), 2, . . . , show that
\(E\left(Z^{2 i}\right)=\frac{2^{i} \Gamma\left(i+\frac{1}{2}\right)}{\sqrt{\pi}}.\)
[Hint: A function \(h(\cdot)\) is an even function if, for all \(y,\ h(-y)=h(y)\). For any even function \(h(y),\ \int_{-\infty}^{\infty}h(y)\ dy=2\int_0^{\infty}h(y)\ dy\) if the integrals exist. Use this fact, make the change of variable \(w=z^{2} / 2\), and use what you know about the gamma function.]
c Use the results in part (b) and in Exercises 4.81(b) and 4.194 to derive \(E\left(Z^2\right),\ E\left(Z^4\right),\ E\left(Z^6\right),\text{ and }E\left(Z^8\right)\).
d If \(i=1\), 2, . . . , show that
\(E\left(Z^{2 i}\right)=\prod_{j=1}^{i}(2 j-1).\)
This implies that the ith even moment is the product of the first 𝑖 odd integers.
Equation Transcription:
Text Transcription:
i=1
E(Z^2i-1)=0
g(cdot)
y,g(-y)=-g(y)
g(y), integral -infinity to infinity g(y) dy=0
i=1
E(Z^2i)=2^i Gamma(i+1over 2)over sqrt pi.
h(cdot)
y,h(-y)=h(y)
h(y), integral -infinity to infinity h(y) dy=2 integral 0 to infinity h(y) dy
w=z^2/2
E(Z^2), E(Z^4), E(Z^6), and E(Z^8)
i=1
E(Z^2i)=product j=1 to i (2j-1).
ANSWER:
Answer:
Step 1 of 4:
(a)
Let be a standard normal variable.
We need to show that expected values of all odd integer powers of are
That is if show that
A standard normally distributed random variable with parameters (0, 1), i.e. has density therefore,
……(1)
Now,
We know for any odd function we can write,
Let define function
Lets check the properties of odd function.
hence the is an odd function.
Since the is an odd function, we can say that
is also an odd function.
Hence from the property of odd function, we can write,
Hence proved.