MHB Greatest probability - Expected value

[mathmari]

Hey!

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? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

The value with the greatest probably is known as the Mode, which is:
$$\text{Mode} = \mathop{\mathrm{arg\,max}}_{x\in \mathbb N} f_X(x)$$
It is one of the Center metrics, just like Mean and Median.
However, the Expected Value, also known as Mean, is the average weighted on probability, or:
$$\text{Expected Value} = \sum_{x\in\mathbb N}xf_X(x)$$
If the distribution is symmetric, they are the same, but otherwise they are not. (Thinking)

 

[mathmari]

The value with the greatest probably is known as the Mode, which is:
$$\text{Mode} = \mathop{\mathrm{arg\,max}}_{x\in \mathbb N} f_X(x)$$

What do you mean by arg? I got stuck right now. (Wondering)

However, the Expected Value, also known as Mean, is the average weighted on probability, or:
$$\text{Expected Value} = \sum_{x\in\mathbb N}xf_X(x)$$
If the distribution is symmetric, they are the same, but otherwise they are not. (Thinking)

Ah ok!

 

[I like Serena]

What do you mean by arg? I got stuck right now.

I introduced the $\mathrm{arg\,max}$ notation only to illustrate the difference with the expected value.
$\mathrm{arg\,max}$ is the value (the argument) for which the given expression takes its maximum. (Nerd)

In this case we can calculate the expected value with:
$$\text{Expected Value} = \sum_{x\in\mathbb N}xf_X(x) = \sum x p(1-p)^{x-1}
= p\sum x (1-p)^{x-1} = p \sum \d{}p\Big[-(1-p)^x\Big] \\
= -p \d{}p\left[ \sum (1-p)^x\right] = -p\cdot \d{}p\left[ \frac{1}{1-(1-p)}\right]
= -p\cdot \d{}p\left[ \frac 1p\right] = -p \cdot -\frac 1{p^2} = \frac 1p
$$
(Thinking)