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The final exam solutions for a probability theory course (math/stats 525) with 8 problems covering topics such as joint density functions, expected values, moment generating functions, and poisson distributions.
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Friday, Dec. 16, 2005, 10:40am–12:30am
Instructions:
Problem Possible score Your score 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 Total 80
f (x, y) =
(x + y)e−(x+y), x, y > 0.
Find the density function of Z = X + Y.
We have a wooden stick of length 1. Break the stick at a random point. The left piece has length X which is uniformly distributed on (0, 1). Break this piece further at a random point. The resulting left piece has length Y where Y is uniformly distributed on (0, X). Find (a) EY (b) fY |X (y|x) (c) fY (y).
Let X 1 , X 2 ,... be independent Exp(1) random variables. The density function of Sn = X 1 +· · ·+Xn is known to be fSn (z) =
zn−^1 e−z (n − 1)!
, z > 0
for n = 1, 2 ,.... Here 0! = 1 as usual. You don’t need to prove this formula. Now fix λ > 0. Let N = min{n : X 1 + X 2 + · · · + Xn+1 > λ}. Find the distribution of N using the density function of Sn above.
(c) Using the result of (a), compute the moments E(Xk) of X ∼ N (0, 1) for all k = 1, 2 , 3 ,....
(a) Show that the number X of heads and the number Y of tails are independent.
(b) Find the distribution of X.
(a) Let S = X + Y where X ∼ P ois(λ) and Y ∼ P ois(μ) are independent random variables. Show that S ∼ P ois(λ + μ). (Hint: You can use either the convolution formula or the probability generating function. The probability generating function of X ∼ P ois(λ) is E(zX^ ) = eλ(z−1).)
(b) Let S = X 1 +... Xn where X 1 , X 2 ,... , Xn are independent P ois(1) random variables. Show that S ∼ P ois(n).
(c) Let Sn ∼ P ois(n) where n is a positive integer. Prove that for each x,
nlim→∞ P
Sn − n √ n
≤ x
∫ (^) x
−∞
2 π
e−^
(^12) y 2 dy.
Facts that may be useful:
∑^ ∞
k=
xk^ =
1 − x
, |x| < 1
k=
xk k!
= ex
k=
(−1)k−^1 k
xk^ = log(1 + x), |x| < 1
(x + y)n^ =
∑^ n
k=
n k
xkyn−k
n^ lim→∞
α n
)n = eα
Some important distributions:
(n k
pk(1 − p)n−k, k = 0,... , n EX = np, varX = np(1 − p)
k k! ,^ k^ ∈ {^0 ,^1 ,^2 ,.. .}^ EX^ =^ λ, varX^ =^ λ,^ G(z) =^ E(z
X (^) ) = eλ(z−1)
2
−(x−μ)
2 2 σ^2
EX = μ, varX = σ^2.
