Question

(25 pts) Consider a random variable X with density function f X (x) = x , 0 ≤ x ≤1 0, else ⎧ ⎨ ⎪ ⎩ ⎪ . a. (3) Carefully sketch by hand the density function f X (x) . Be sure to dimension both axes. b. (4) Find the mean and variance of X . c. (4) Suppose we generate a new random variable Y = Q(X) by quantizing X according to the following 3-level uniform quantizer: Q(x) = − 2 3 , −1≤ x < − 1 3 0, − 1 3 ≤ x < 1 3 2 3 , 1 3 ≤ x ≤1

Answer 1

Answer:

Step-by-step explanation:

It seems that this question is incomplete, and unfortunately, no reference was found in the internet. However, it seems that question is about calculating the variance and the mean of a random variable, based on its pdf (probability density funtion). Recall that a function must fulfill the following property for it to be a pdf

$\int_{-\infty}^{\infty} f(x) dx =1$

Also, recall that the following formulas

$\text{E}\[X\] = \int_{-\infty}^{\infty} xf(x) dx$ (the mean)

$\text{Var}\[X\] = \int_{-\infty}^{\infty} x^2f(x) dx-(\text{E}\[X\])^2$.

Let us illustrate this calculations with an example.

Consider the function f(x) = 2x if $0\leq x \leq 1$ and 0 otherwise. By easy calculations, we can check that f(x) is indeed a pdf (it integrates up to 1). Hence it's mean is

$\text{E}\[X\] = \int_{0}^{1} x\cdot2x dx = \left.\frac{2}{3}x^3\rigth|_{0}^1 = \frac{2}{3}$

and the variance is given by

$\text{Var}\[X\] = \int_{0}^{1} x^2\cdot 2x dx -(\frac{2}{3})^2= \left.\frac{1}{2}x^4\right|_{0}^1-(\frac{2}{3})^2 = \frac{1}{2}-\frac{4}{9}= \frac{1}{18}$