Question

Problem 31. Calculate the expected value and variance of X for each of the following scenarios.

1. X = {0, 1} where each has equal probability. (A coin flip)

2. X = {1, 2, 3, 4, 5, 6} where each has equal probability. (A die roll)

3. X = {0, 1} with f(0) = 1/3 and f(1) = 2/3.

4. X = B(3, 0.35). (Use info from Problem 26.)(Problem 26. Let X = B(3, 0.35). Calculate each f(k) and the sum X 3 k=0 f(k).)

Problem 32. Calculate the expected value and variance of X = B(3, 0.35) by using Theorem 41. Compare the results to part 4 of Problem 31.

Problem 33. Let (X, f) be a CPD. Show that P(X = x) = 0 for any x ∈ X.

Problem 34. Consider f : R → R defined by f(x) = 1 1 + x 2 . Explain why f is not a PDF, and find a constant c so that cf is a PDF.

Problem 35. Let F be a CDF for a CPD (X, f). Find lim x→−∞ F(x) and limx→∞ F(x).

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Answer 1

Problem 31.

1. X = {0, 1} where each has equal probability. (A coin flip)

For fair coin flip, the probability distribution is,

f(X = 0) = f(X = 1) = 1/2

Expected value of X is,

E(X) = 0 * f(X = 0) + 1 *  f(X = 1) = 0 * (1/2) + 1 * (1/2) = 1/2

E(X²) = 0² * f(X = 0) + 1² *  f(X = 1) = 0² * (1/2) + 1² * (1/2) = 1/2

Variance of X is,

Var(X) = E(X²) - E(X)²   = (1/2) - (1/2)² = (1/2) - (1/4) = 1/4

2. X = {1, 2, 3, 4, 5, 6} where each has equal probability. (A die roll)

For fair dice roll, the probability distribution is,

f(X = 1) = f(X = 2) = (X = 3) = f(X = 4) = (X = 5) = f(X = 6) = 1/6

Expected value of X is,

E(X) = 1 * f(X = 1) + 2 * f(X = 2) + 3 * f(X = 3) + 4 * f(X = 4) + 5 * f(X = 5) + 6 * f(X = 6)

= (1/6) * 1 + (1/6) * 2 + (1/6) * 3 + (1/6) * 4 + (1/6) * 5 + (1/6) * 6

= 3.5

E(X²) = 1² * f(X = 1) + 2² * f(X = 2) + 3² * f(X = 3) + 4² * f(X = 4) + 5² * f(X = 5) + 6² * f(X = 6)

= (1/6) * 1 + (1/6) * 4  + (1/6) * 9 + (1/6) * 16 + (1/6) * 25 + (1/6) * 36

= 15.16667

Variance of X is,

Var(X) = E(X²) - E(X)²   = 15.16667 - (3.5)² =2.91667

3. X = {0, 1} with f(0) = 1/3 and f(1) = 2/3.

Expected value of X is,

E(X) = 0 * f(X = 0) + 1 *  f(X = 1) = 0 * (1/3) + 1 * (2/3) = 2/3

E(X²) = 0² * f(X = 0) + 1² *  f(X = 1) = 0² * (1/3) + 1² * (2/3) = 2/3

Variance of X is,

Var(X) = E(X²) - E(X)²   = (2/3) - (2/3)² = (2/3) - (4/9) = 2/9

4. X = B(3, 0.35).

n = 3, p = 0.35

Expected value of X is,

E(X) = np = 3 * 0.35 = 1.05

Variance of X is,

Var(X) = np(1-p) = 3 * 0.35 * (1 - 0.35) = 0.6825