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Firepanda
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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
thanks
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...
Firepanda said:Sorry I should have stated before that I don't know where to start on this.
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.
No, only 1 to n, the number over which your probability distribution is defined.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?
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?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?
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.
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.
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.
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.
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.