Find c for Random Variable: E(X), E(X^2), E(1/X) & Var(X)

In summary: Otherwise you have the right idea. c doesn't have to equal infinity. what is the sum of n consecutive integers?The sum of the first n positive integers is n.
  • #1
Firepanda
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0
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
 
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  • #2
What is your solution?

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

The expectations are obtained directly by definition...
 
  • #3
maverick280857 said:
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
 
  • #4
Firepanda said:
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.

Firepanda said:
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?
 
  • #5
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?
 
  • #6
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?
 
  • #7
Firepanda said:
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?
 

What is the expected value (E) of a random variable X?

The expected value of a random variable X is the average value that the variable takes on over many trials or observations. It is calculated by summing the products of each outcome of X and its corresponding probability.

What is E(X^2) for a random variable X?

E(X^2) is the expected value of the squared values of X. It is calculated by summing the products of each squared outcome of X and its corresponding probability.

How do you find the constant (c) for E(1/X) for a random variable X?

To find the constant (c) for E(1/X), you can use the formula E(1/X) = 1/c, where c is the constant. This can also be written as E(X^-1) = 1/c. You can then solve for c by setting the expected value equal to its given value and solving for c.

What is the variance (Var) of a random variable X?

The variance of a random variable X is a measure of how much the values of X vary from the expected value. It is calculated by finding the expected value of (X-E(X))^2, where E(X) is the expected value of X.

How do you find c for a random variable X given its expected value and variance?

To find c for a random variable X, you can use the formula c = E(X)/Var(X), where E(X) is the expected value of X and Var(X) is the variance of X. This formula can also be written as c = E(X)/E[(X-E(X))^2]. You can then plug in the values for E(X) and Var(X) to solve for c.

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