Question

Let RV x have the pro dist f(x)=q^x-1.pi x=1, 2, 3... Find the mgf of x and hence its mean and variance?

Answer 1

Sum of probabilities $\sum_{i=1}^\infty f(x_i)=1$

Therefor $\sum_{i=1}^\infty q^{x_i-1}\cdot p=$

$=p\sum_{i=1}^\infty q^{i-1}=\\ p\sum_{i=0}^\infty q^i=\frac{p}{1-q}=1$ and $p=1-q$

Moment generating function

$M_x(t)=\displaystyle\sum_{i=1}^\infty e^{tx_i}\cdot f(x_i)=$

$\displaystyle\sum_{i=1}^\infty e^{t\cdot i}q^{i-1}\cdot p=p/q\cdot\displaystyle\sum_{i=1}^\infty e^{t\cdot i}q^{i}=\\ p/q\cdot\displaystyle\sum_{i=1}^\infty (e^tq)^i$

Sum of geometric series $\sum_{i=1}^\infty (e^tq)^i=e^tq\cdot\frac{1}{1-e^tq}$

$M_x(t)=p/q\frac{e^tq}{1-e^tq}=\frac{pe^t}{1-qe^t}$

Expectations

$E(x)=M_x'(0)=(\frac{pe^t}{1-qe^t})'_{t=0}$

$E(x^2)=M_x''(0)=(\frac{pe^t}{1-qe^t})''_{t=0}$

$(\frac{pe^t}{1-qe^t})'=\frac{pe^t(1-qe^t)-pe^t(-qe^t)}{(1-qe^t)^2}= \frac{pe^t}{(1-qe^t)^2}$

$(\frac{pe^t}{1-qe^t})''=(\frac{pe^t}{(1-qe^t)^2})'= \frac{pe^t(1-qe^t)^2-pe^t\cdot2(1-qe^t)(-qe^t)}{(1-qe^t)^4}=\\ pe^t\frac{1-qe^t+2qe^t}{(1-qe^t)^3}=pe^t\frac{1+qe^t}{(1-qe^t)^3}$

$E(x)=(\frac{pe^t}{(1-qe^t)^2})_{t=0}=\frac{p}{(1-q)^2}=p/p^2=1/p$

$E(x^2)=(pe^t\frac{1+qe^t}{(1-qe^t)^3})_{t=0}=p\frac{1+q}{(1-q)^3}=\frac{1+q}{p^2}$

Variance of x

$Var(x)=E(x^2)-(E(x))^2=\frac{1+q}{p^2}-\frac{1}{p^2}=q/p^2=\\ \frac{1-p}{p^2}$

Answer: $E(x)=1/p,\ Var(x)=\frac{1-p}{p^2}.$