Expected Value of Random Variables

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To calculate the expected value E(X) from the given cumulative distribution function (cdf), one must first derive the probability mass function (pmf) from the cdf. The discontinuous jumps in the cdf indicate the probabilities at specific values of the random variable. The pmf is determined to be 0.25 for x = -1, 0.25 for x = 0, 0.25 for x = 1, and 0.25 for x = 4. With the pmf established, E(X) can be calculated using the formula E(X) = x1*p(x1) + x2*p(x2) + ... for the respective values. This approach effectively allows for the computation of the expected value from the cdf.
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Homework Statement


Consider a random variable X having cdf:

1, x ≥ 4,
3/4, 1 ≤ x < 4,
FX(x) = 1/2, 0 ≤ x < 1,
1/4, −1 ≤ x < 0,
0, x < −1.

Give the value of E(X).

Homework Equations





The Attempt at a Solution


I know how to calculate the value of E(X) given the probability mass function (E(X) = x1*p(x1) + x2*p(x2) + ...) but how do I calculate E(X) given the cdf?
 
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You can find the probability density (what you are referring to as the probability mass function) from the cdf. Think about what the discontinuous jumps in the cdf represent.
 
Hmm.. so would the pmf be:

0.25, x = -1
0.25, x = 0
0.25, x = 1
0.25, x = 4

?
 
Yup!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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