Question

Suppose we generate a random variable X in the following way. Forst we flip a fair coin. If the coin is heads, take X to have a Uniform(0, 1). If the coin is tails, take X to have a Uniform(3, 4). (a) Find the mean of X. (b) Find the standard deviation of X.

Answer 1

Answer:

(a) The mean of X is 2.

(b) The variance X is $\frac{7}{3}$.

Step-by-step explanation:

The random variable X defined as follows:

X = heads, then X follows Uniform (0, 1).

X = tails, then X follows Uniform (3, 4).

(a)

The expected value of a Uniform random variable U (a, b) is:

$Mean=\frac{1}{2}(a+b)$

Compute the expected value of X = heads as follows:

$E(X=heads)=\frac{1}{2}(0+1)=\frac{1}{2}=0.50$

Compute the expected value of X = tails as follows:

$E(X=tails)=\frac{1}{2}(3+4)=\frac{7}{2}=3.50$

Compute the mean of X as follows:

$E(X)=\frac{E(X=heads)+E(X=tails)}{2} =\frac{0.50+3.50}{2}=\frac{4.00}{2}=2$

Thus, the mean of X is 2.

(b)

Compute the variance of X as follows:

$V(X)=\frac{1}{2}\int\limits^{1}_{0} {\frac{1}{1-0}[X-E(X)]^{2}} \, dx +\frac{1}{2}\int\limits^{4}_{3} {\frac{1}{4-3}[X-E(X)]^{2}} \, dx \\=\frac{1}{2}\int\limits^{1}_{0} {[x-2]^{2}} \, dx +\frac{1}{2}\int\limits^{4}_{3} {[x-2]^{2}} \, dx \\=\frac{1}{2}[\frac{(x-2)^{3}}{3}]^{1}_{0}+\frac{1}{2}[\frac{(x-2)^{3}}{3}]^{4}_{3}\\=(\frac{1}{2}\times\frac{7}{3})+(\frac{1}{2}\times\frac{7}{3})\\=\frac{7}{3}$

Thus, the variance X is $\frac{7}{3}$.