Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What would be the expected value of a random variable with distribution $\displaystyle \frac{1}{k^2-k}$? I'm basically stuck. $k \in [2, \infty) \cap \mathbb N$.

share|improve this question
1  
What are the values? $k \in \mathbb{N}$? –  vonbrand Mar 18 '13 at 0:49
    
It should be $k\in[2,\infty)\cap\mathbb{N}$ I think because otherwise the integral does not sum up to 1. –  NECing Mar 18 '13 at 0:55
    
@ShuXiaoLi: I think you are right and updated problem. –  Amzoti Mar 18 '13 at 1:05

3 Answers 3

up vote 4 down vote accepted

Firstly, let's be a little more careful with our wording. Assuming $X$ is a random variable with probability mass function $$ p(k) = \frac{1}{k^2-k} = \frac{1}{k(k-1)} = \frac{1}{k-1}-\frac{1}{k} $$ for $k \geq 2$, we would like to calculate $E[X]$.

First note that $\sum_{k=1}^\infty p(k)$ is nonnegative a telescoping series which sums to $1$, so it indeed defines a valid probability measure.

Then we calculate $$ E[X] = \sum_{k=2}^\infty k p(k) = \sum_{k=2}^\infty \frac{1}{k-1} = \infty. $$

share|improve this answer
    
So the fact that it goes off to infinity does not necessarily mean that I calculated the distribution incorrectly? –  John Mar 18 '13 at 0:56
    
No, it does not mean that you did anything wrong. Some random variables have infinite expectation (including this one). One should be careful though, if the random variable is allowed to take negative values then $E[X]$ might not be well defined. If $X \geq 0$ (like this case) you need not worry though. –  nullUser Mar 18 '13 at 0:59
    
How does one arrive at the fact that this summation goes off to infinity? From ∑k=2->∞ 1/k−1 to this: ∞. –  John Mar 18 '13 at 22:44
    
This is the harmonic series which is well known to diverge. en.wikipedia.org/wiki/Harmonic_series_(mathematics) There are many proofs that can be found on wikipedia for this result. –  nullUser Mar 18 '13 at 23:23

If $p_k = \dfrac{1}{k^2 - k}$, then: $$ \sum_{k \ge 2} \frac{1}{k^2 - k} = \sum_{k \ge 2} \left( \frac{1}{k - 1} - \frac{1}{k} \right) = 1 $$ OK, so we want: $$ \mathbb{E}(k) = \sum_{k \ge 2} \frac{k}{k^2 - k} = \sum_{k \ge 2} \frac{1}{k - 1} = \sum_{n \ge 1} \frac{1}{n} $$ This sum doesn't exist (the harmonic series diverges).

share|improve this answer

$$\mathbb{E}(X)=\sum_{k=2}^{\infty}\frac{k}{k^2-k}=\sum_{k=2}^{\infty}\frac{1}{k-1}=\infty$$ Does not converge.

share|improve this answer
    
That's what I thought. If this is the case, does that most likely mean that I am approaching the problem incorrectly? –  John Mar 18 '13 at 0:54
    
@John There are many distributions that do not have a "mean". So ending up with divergent mean does not mean incorrect. –  NECing Mar 18 '13 at 0:57
    
@John, not necessarily. The result is OK, the probabilities sum up to 1 as they should. –  vonbrand Mar 18 '13 at 0:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.