- Thread starter
- #1

- Apr 14, 2013

- 4,036

The geometric distribution with parameter $p\in (0,1)$ has the probability function \begin{equation*}f_X(x)=p(1-p)^{x-1}, \ \ x=1, 2, 3, \ldots\end{equation*}

I have shown that $f_X$ for each value of $p\in (0,1)$ is strictly monotone decreasing, as follows:

\begin{align*}f_X(x+1)=p(1-p)^{x+1-1}=p(1-p)^{(x-1)+1}=p(1-p)^{x-1}(1-p)\overset{(\star)}{<}p(1-p)^{x-1}=f_X(x)\end{align*} $(\star)$ : Since $p\in (0,1)$ we have that \begin{equation*}0<p<1\Rightarrow -1<-p<0 \Rightarrow 0<1-p<1\end{equation*}

That means that the value $x=1$ has the greatest probability. But the expected value of a random variable $X$ with geometric distribution is $\frac{1}{p}$. Why is it like that and not equal to the value with the greatest probability?