A pdf is given up to a multiplicative factor, find it, then find the cdf, and some of its moments

Question

Let $X$ be random variable with mass function (p.m.f): $f(x)=c(\frac{4}{9})^x$ ; $x=1,2,3,...$

a. determine the value of C that makes f p.m.f

b. determine the distribution function of X

c. calculate the mean and variance of X

Answer 1

For the mean, which is

$$c\sum_{k=1}^{\infty}x(\tfrac49)^x=c.\left(1(\tfrac49)^1+2(\tfrac49)^2+3(\tfrac49)^3+4(\tfrac49)^4+\cdots \right),$$

there is a classical trick : rewrite this sum in columns under the form:

$$c.\left\{\begin{array}{cccccc} (\tfrac49)^1&+(\tfrac49)^2&+(\tfrac49)^3&+(\tfrac49)^4&+(\tfrac49)^5&\cdots \\ &+(\tfrac49)^2&+(\tfrac49)^3&+(\tfrac49)^4&+(\tfrac49)^5&\cdots \\ &&+(\tfrac49)^3&+(\tfrac49)^4&+(\tfrac49)^5&\cdots \\ &&&+(\tfrac49)^4&+(\tfrac49)^5&\cdots \\ &&&&+(\tfrac49)^5& \cdots \end{array}\right.$$

Now, let us sum it by beginning by summations of its lines.

As line number $m$ can be written under the form:

$$c.(\tfrac49)^m\sum_{x=0}^{\infty}(\tfrac49)^x=c.(\tfrac49)^m\dfrac{1}{1-\tfrac49}=c.(\tfrac49)^m\tfrac95$$

(sum of a geometric series beginning at exponent $0$).

Summing all these sums, one gets :

$$c.\tfrac95\sum_{m=1}^{\infty}(\tfrac49)^m=c.\tfrac95.\tfrac49.\sum_{m=0}^{\infty}(\tfrac49)^m=c.\tfrac95.\tfrac49.\tfrac95.$$

(we have begun by the factorization of $\tfrac49$ in order to have a geometric series beginning at $m=0$, whose sum, as before, is $\tfrac95$).

Thus the mean is $c.\tfrac{36}{25}.$