Find the mean and second moment of the random variable whose PDF has the characteristic function ΦX(w) = exp(jwμX - w²σ²X/2)

Question

Find the mean and second moment of the random variable whose PDF has the characteristic function

[latex]\,\displaystyle \Phi_X\left (w\right )=\exp \left (j w \mu_X-\frac{w^2 \sigma_X^2}{2}\right )[/latex]

Accepted Answer

The first and second derivatives of [latex]\Phi_X(w)[/latex] are given by

[latex]\,\displaystyle \begin{aligned} \frac{d}{d w} \Phi_X\left (w ight ) & =\left (j \mu_X-w \sigma_x^2 ight ) \exp \left (j w \mu_X-\frac{w^2 \sigma_X^2}{2} ight ) \ \frac{d^2}{d w^2} \Phi_X\left (w ight ) & =\left\{\left (j \mu_X-w \sigma_X^2 ight )^2-\sigma_x^2 ight\} \exp \left (j w ~\mu_X-\frac{w^2 \sigma_X^2}{2} ight ) \end{aligned}[/latex]

Thus, we obtain

[latex]\,\displaystyle \begin{aligned} E[X] & =\left.\frac{1}{j} \frac{d}{d w} \Phi_X\left (w ight ) ight\vert _{w=0}=\left.\left (j \mu_X-w \sigma_x^2 ight ) \exp \left (j w ~ \mu_X-\frac{w^2 \sigma_X^2}{2} ight ) ight\vert _{w=0}=\mu_X \ E\bigg[X^2\bigg] & =-\left.\frac{d^2}{d w^2} \Phi_X\left (w ight ) ight\vert _{w=0}=-\left.\left\{\left (j \mu_X-w \sigma_x^2 ight )^2-\sigma_x^2 ight\} \exp \left (j w~ \mu_X-\frac{w^2 \sigma_X^2}{2} ight ) ight\vert _{w=0} \ & =\mu_X^2+\sigma_x^2 \end{aligned}[/latex]

Question 7.1: Find the mean and second moment of the random variable whose......

Related Answered Questions

Verified Answer:

Verified Answer:

Verified Answer:

The z-transform of the PMF of a discrete random variable K is given by GK(z) = A[10 + 8z²/2 – z] a. What is the expected value of K? b. Find , the probability that K has the value 1. ...

Verified Answer:

Explain why the function F(z) = z² + z – 1 is or is not a valid z-transform of the PMF of a random variable. ...

Verified Answer:

Is the following function a valid z-transform of a PMF? If so, what is the PMF? g(z) = 1 – a/1 – az 0 < a < 1 ...

Verified Answer:

Determine in an efficient manner the fourth moment of the random variable X whose PDF is given by fX(x) = 32x²e^-4x x ≥ 0 ...

Verified Answer:

A communication channel is degraded beyond use in a random manner. A smart student figured out that the duration Y of the intervals between consecutive periods of degradation has the PDF fY(y) = {0.2^4y³e^-0.2y/3! y ≥ 0 What is the s-transform of the PDF of Y? ...

Verified Answer:

For what value of K is the function A(s) = K/(s + 5) a valid s-transform of a PDF? ...

Verified Answer: