# Random variables

a discrete random variable has range space {1, 2, ..., n} and satisfies P(X=j) = j/c for some number c. Find c, and then find E(X), E(X^2), E(1/X) and Var(X).

thanks

## Answers and Replies

What is your solution?

You have the probability mass function (pmf). You can determine the constant c.

The expectations are obtained directly by definition...

What is your solution?

You have the probability mass function (pmf). You can determine the constant c.

The expectations are obtained directly by definition...

Thanks for your reply.

Sorry I should have stated before that I don't know where to start on this.

I just looked up on pmf, and I have no examples on a question like this in my notes.

The 'j' confuses me in the question as I don't see how it relates to anything else, so finding c is tricky for me.

Thanks

quadraphonics
Sorry I should have stated before that I don't know where to start on this.

Well, aren't there a few conditions that all valid pmf's are required to satisfy? It would probably be a good idea to review these.

The 'j' confuses me in the question as I don't see how it relates to anything else, so finding c is tricky for me.

j is simply a dummy variable. Just shorthand for saying P(X=1) = 1/c, P(X=2) = 2/c, ..., P(X=n) = n/c.

Perhaps this thread should be moved to the homework help section?

The conditions I know of pmf are that the total sum of the probabilities from -infinity to infinity is 1, and the probabilities can only take values between 0 and 1.

So do I have to find a j/c which has a sum of the series from -infinity to infinity equal to 1?

Sorry I'm really confused..

ok so I have so far:

1/c + 2/c + 3/c ... +n/c = 1

1 + 2 + 3 +.. + n = c

c = infinity?

a pmf is for a discrete random variable.

Do you know what the definition of a discrete RV is?

Otherwise you have the right idea. c doesn't have to equal infinity. what is the sum of n consecutive integers?

HallsofIvy
Science Advisor
Homework Helper
The conditions I know of pmf are that the total sum of the probabilities from -infinity to infinity is 1, and the probabilities can only take values between 0 and 1.

So do I have to find a j/c which has a sum of the series from -infinity to infinity equal to 1?
No, only 1 to n, the number over which your probability distribution is defined.

Sorry I'm really confused..

ok so I have so far:

1/c + 2/c + 3/c ... +n/c = 1

1 + 2 + 3 +.. + n = c

c = infinity?
Why should it be? n is a fixed finite number, not "infinity". Do you know the formula for the sum of the first n positive integers?