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Greatest probability - Expected value

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Hey!! :eek:

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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,003
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!!
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
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)