Question

Given that p(x) =k÷x² is a probability distribution for a random variable that can take on the values x = 0, 1, 2, 3, and 4.

a. Find k.

b. Find the expression for the cdf F(x) of the random variable. 

Answer 1

a. Finding K

$\sum_{i=0}^4P(Xi)=1$

The PMF should be $\frac{k}{2^x}$ and not $\frac{k}{x^2}$ since we cannot divide a number by zero.

$\sum_{i=0}^4P(Xi)=k+\frac{k}{2}+\frac{k}{4}+\frac{k}{8}+\frac{k}{16}=\frac{31k}{16}$

$\frac{31k}{16}=1$ Solving for k,

$k=\frac{16}{31}$

b. CDF

$CDF=\sum_0^xP(x)$ Substituting the value of k in $\sum_{i=0}^4P(Xi)$ above, we get;

$\sum_0^4P(x)=\frac{16}{31}+\frac{8}{31}+\frac{4}{31}+\frac{2}{31}+\frac{1}{31}$

Note that it forms a geometric progression with $a=\frac{16}{31}$ and $r=\frac{1}{2}$. Note also that it consist of 5 terms since the value of x starts from 0.

The sum of n terms of a GP when $r<1$ is

$S_n=\frac{a(1-r^n)}{1-r}$

But to allow 0 to 4 values of x, n becomes x+1.

$CDF=\frac{\frac{16}{31}(1-\frac{1}{2^{x+1}})}{1-\frac{1}{2}}$ Simplifying, we have

$CDF=\frac{32}{31}(1-\frac{1}{2^{x+1}})$