# Expected value of probability distribution

• Jun 4th 2010, 10:34 PM
acevipa
Expected value of probability distribution
The number $\displaystyle X$ of particles emitted as the result of an experiment is a random variable with probability distribution:

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

What is the expected number of particles emitted during one experiment

This is how I've gone so far:

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

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

I'm confused as to how to evaluate this sum
• Jun 5th 2010, 02:49 AM
mr fantastic
Quote:

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

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

What is the expected number of particles emitted during one experiment

This is how I've gone so far:

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

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

I'm confused as to how to evaluate this sum

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

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

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

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

$\displaystyle \Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = r^2(....)$
• Jun 5th 2010, 02:55 AM
SpringFan25
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:

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

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

edit too slow :D
• Jun 5th 2010, 03:05 AM
SpringFan25
oops i had the wrong form of the geometric distribution.

its $\displaystyle \frac{1-p}{p}$
• Jun 5th 2010, 03:06 AM
acevipa
Quote:

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

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

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

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

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

Yes, but for my case,

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

So if you differentiate that, it just becomes 0
• Jun 5th 2010, 03:11 AM
mr fantastic
Quote:

Originally Posted by acevipa
Yes, but for my case,

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

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.
• Jun 5th 2010, 03:34 AM
acevipa
Ok, so is this correct:

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

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

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

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

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

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

$\displaystyle =1$
• Jun 5th 2010, 03:42 AM
mr fantastic
Quote:

Originally Posted by acevipa
Ok, so is this correct:

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

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

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

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

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

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

$\displaystyle =1$

Yes. (Verification is simple using the well known formula given here: Geometric distribution - Wikipedia, the free encyclopedia).
• Jun 5th 2010, 03:50 AM
acevipa
Quote:

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.