Question

Probability mass function and p-series:

Answer 1

From calculus it is known that the series \[\sum_{}^{}\] \[1\div n^{2}\] converges. Let c denote the sum of this series. It follows that the function p(n) = 1/(cn)^2, n = 1, 2, ..., defines a probability mass function. Show that the expected value of a random variable with this probability mass function does not exist. (i.e., is infinite).

Answer 2

Are you sure the c is squared?

Answer 3

In any case, you need to evaluate the expected value, which is by definition:
\[\sum_{n=1}^{\infty}n p(n)\]

Answer 4

the c is not squared,, sorry. That should be 1/cn^2

Answer 5

So does \[\sum_{n=1}^{\infty}np(n)\] converge?

Answer 6

I don't know. I know that the series converges only for values of the exponent > 1, but it diverges for values of the exponent between 0 and 1. So, does this prob mass function not exist because the marginal probabilities themselves have to sum to 1, but the series itself does not converge for values between 0 and 1. I'm so confused.

Answer 7

Let's see, the expected value=

\[\sum_{n=1}^{\infty}np(n)=\sum_{n=1}^{\infty}\frac{1}{cn}=\frac{1}{c}\sum_{n=1}^{\infty}\frac{1}{n}\]

Answer 8

You were this far?

Answer 9

yes,, the n is squared,, 1/cn^2

Answer 10

Because we are calculating an expected value we multiply by n, so the n squared will become just n:

np(n)=n/(cn^2)=1/(cn)

Are you familiar with the definition of the expected value? So multiplying by n and taking the sum?

Answer 11

I'm just now learning the section over expected value. It doesn't seem difficult, but I think that putting together the probability mass functions is what's stumping me the most. I understood other expected value problems in the chapter, but these had the prob mass functions already defined for me.

Answer 12

But the probability mass function is given here too: p(n)=1/cn^2

Answer 13

let me get my tablet so I can write...brb

Answer 14

and I don't understand what's the deal with the "c",, I'm never going to have a sum!

Answer 15

my expected value calculation would look something like: 1(1) + 2(1/2) + 3(1/3)....right?

Answer 16

About the c:
So c is defined as \[c=\sum_{n=1}^{\infty}\frac{1}{n^2}\].
If c isn't equal to one, than this isn't a probability mass function, right, because the sum isn't equal to 1. If we divide each term by c though, the sum is equal to 1.

Answer 17

the expected value would be this: 1(1/c)+2(1/4c)+3(1/9c)+...

Answer 18

So I'm not sure what your problem is, do you understand the notation for series:\[\sum_{n=1}^{\infty}\]

Answer 19

Yes, I understand series....the different types of them in addition to the different types of tests to tell whether or not they converge or diverge.

Answer 20

I'll move onto some other problems and chew on the info you've provided me for this problem later on.

Answer 21

Thanks for your help

Answer 22

OK, if you have more questions about this, you can post them here, I'll probably see them eventually.

Answer 23

ok, thanks