# Thread: Expected value of probability distribution

1. ## Expected value of probability distribution

The number $X$ of particles emitted as the result of an experiment is a random variable with probability distribution:

$P(X=k)=\left(\frac{1}{2}\right)^{k+1}\ for\ k \geq 0$

What is the expected number of particles emitted during one experiment

This is how I've gone so far:

$E(X)=\sum_{k=0}^{\infty}kP(X=k)$

$=\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\rig ht)^3+ 3\left(\frac{1}{2}\right)^4+...$

I'm confused as to how to evaluate this sum

2. Originally Posted by acevipa
The number $X$ of particles emitted as the result of an experiment is a random variable with probability distribution:

$P(X=k)=\left(\frac{1}{2}\right)^{k+1}\ for\ k \geq 0$

What is the expected number of particles emitted during one experiment

This is how I've gone so far:

$E(X)=\sum_{k=0}^{\infty}kP(X=k)$

$=\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\rig ht)^3+ 3\left(\frac{1}{2}\right)^4+...$

I'm confused as to how to evaluate this sum
It is well known that the sum of an infinite geometric series is $\sum_{k=0}^{+\infty} r^k = \frac{1}{1-r}$ for |r| < 1. Therefore:

$\frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = ....$

$\Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = ....$

$\Rightarrow \sum_{k=0}^{+\infty}k r^{k-1} = ....$

$\Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = r^2(....)$

3. im pretty rusty on infinite sums and it would be interesting to know if there is a closed form summation of that.

Nevertheless, you dont need to do anything difficult to solve this problem. You can rewrite as follows:

$P(x=k) = 0.5^k \times 0.5$

This is the PF of a geometric distribution with p=0.5. The mean of a geometric distribution is a standard result: $\frac{1}{p}$

edit too slow

4. oops i had the wrong form of the geometric distribution.

its $\frac{1-p}{p}$

5. Originally Posted by mr fantastic
It is well known that the sum of an infinite geometric series is $\sum_{k=0}^{+\infty} r^k = \frac{1}{1-r}$ for |r| < 1. Therefore:

$\frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = ....$

$\Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = ....$

$\Rightarrow \sum_{k=0}^{+\infty}k r^{k-1} = ....$

$\Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = r^2(....)$
Yes, but for my case,

$\left(\sum_{k=0}^{+\infty} r^k\right) = 2$

So if you differentiate that, it just becomes 0

6. Originally Posted by acevipa
Yes, but for my case,

$\left(\sum_{k=0}^{+\infty} r^k\right) = 2$

So if you differentiate that, it just becomes 0
No, the starting point is the first line of my previous reply. Yuo are trying to find a general formula. The, after you find it, you substitute r = 1/2.

7. Ok, so is this correct:

$\frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = \frac{d}{dr} \left(\frac{1}{1-r}\right)$ for $|r|<1$.

$=\frac{1}{(1-r)^2}$

$\Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) =
\sum_{k=0}^{+\infty}k r^{k-1}$

$\Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = \frac{r^2}{(1-r)^2}$ where $r=\frac{1}{2}$

$=\frac{\left(\frac{1}{2}\right)^2}{\left(1-\frac{1}{2}\right)^2}$

$=\frac{\left(\frac{1}{2}\right)^2}{\left(\frac{1}{ 2}\right)^2}$

$=1$

8. Originally Posted by acevipa
Ok, so is this correct:

$\frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = \frac{d}{dr} \left(\frac{1}{1-r}\right)$ for $|r|<1$.

$=\frac{1}{(1-r)^2}$

$\Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) =
\sum_{k=0}^{+\infty}k r^{k-1}$

$\Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = \frac{r^2}{(1-r)^2}$ where $r=\frac{1}{2}$

$=\frac{\left(\frac{1}{2}\right)^2}{\left(1-\frac{1}{2}\right)^2}$

$=\frac{\left(\frac{1}{2}\right)^2}{\left(\frac{1}{ 2}\right)^2}$

$=1$
Yes. (Verification is simple using the well known formula given here: Geometric distribution - Wikipedia, the free encyclopedia).

9. Originally Posted by mr fantastic
Yes. (Verification is simple using the well known formula given here: Geometric distribution - Wikipedia, the free encyclopedia).
Thanks for that.