1. pdf(eX) = C(eF)
Replace X with its respective function
∫-∞∞C(eF) dx = 1
All pdfs must have an integral from negative to positive infinity equal to one
C = 1/(∫-∞∞eF dx)
Use this property to solve for C
pdf(eX) = eF/(∫-∞∞eF dx)
Substitute the value of C
2. E(X) = Σ XP(X) = 0P(X=0) + 1P(X=1) = p
X is a Bernoulli variable, which gives an expected value of p, the probability that x=1.
Var(X) = E(X2) - E(X)2
Start by solving the variance formula
E(X2) = Σ X2P(X) = 0P(X=0) + 1P(X=1) = p
Solve for the expected value of X2
E(X)2 = p2
Use known value of E(X) and square it
Var(X) = p - p2
Substitute values of E(X2) and E(X2)
E(X) = 3Var(X) = p = 3p - 3p2
Set the two sides equal
3p2 = 2p, 3p = 2, p = ⅔
Solve for p
3.
a. αpn ≤ (1-p)pn-1
Substitute α for its maximum. This resembles a geometric distribution with failure
probability (1-p). Since every probability from 1 to ∞ is less than or equal to that
of a geometric distribution, the probability of a family having zero children is ≥p.
b. P(x=k) = Σ[p(1-p)n][(nk)(0.5)n] (sum goes from n=k to ∞)
For every case of k boys, there must be at least k children total. Each number of
kids n has its own probability of occurring, described by a geometric distribution.
For each of these cases, the probability that there are exactly k boys can be
described by a binomial distribution where p=0.5. Multiply these two together and
summate all cases from n=k to n=∞ to find the proportion of families with exactly
k boys.
c. E(n) = ΣnP(n) = Σnp(1-p)n-1 = p[1 + 2(1-p) + 3(1-p)2 + 4(1-p)3 + … + n(1-p)n-1]
Express E(n) in summation form and expand the summation into a series
(1-p)E(n) = p[(1-p) + 2(1-p)2 + 3(1-p)3 + 4(1-p)4 + … + n(1-p)n]
Multiply both sides by (1-p)
E(n) - E(n)(1-p) = p[1 + (1-p) + (1-p)2 + (1-p)3 + … + (1-p)n]
Subtract second sum from first
1 + (1-p) + (1-p)2 + (1-p)3 + … + (1-p)n = 1/p
Simplify the right side to the sum of a geometric series
E(n)(1-(1-p)) = p(1/p)
Substitute
E(n) = 1/p
Simplify
4. (n+1)p[E(1/Y)] = 1 - (1-p)n+1
Substitute X+1 for Y
[E(1/Y)] = Σ[1/(k)](nk)pk(1-p)n-k
Substitute E(Y) for its summation form
Σ(1/k)(nk)pk(1-p)n-k = Σ[P(Y≤k) - P(Y≤k-1)]
Express the summation as a cdf
Σ[P(Y≤k) - P(Y≤k-1)] = P(Y≤n+1) - P(Y≤0)
Simplify cdf by canceling common terms
P(Y≤n+1) - P(Y≤0) = P(X≤n) - P(X≤-1) = 1
Substitute Y for X. A binomial distribution cannot produce a random variable below 0, so
the second term disappears. X will always be less than or equal to n, so the first term is
equal to 1.
(n+1)p[E(1/X+1)] = (n+1)p
Recall what we were solving.
1 - (1-p)n+1 = 1 - (1-p)(1-p)n+1 = 1-(1-p) + p(1-p)n = p[1+(1-p)n]
Simplify the right side of the original equation algebraically.
(n+1)p = p[1+(1-p)n]
(n+1) = 1+(1-p)n
n = (1-p)n
Simplify. I got this one wrong
5. E[1/(X+1)] = Σ[1/(X+1)]p(1-p)n-1
Express E(whatever) as a summation
I give up