Mathematics 375 – Probability and Statistics I Problem Set 5 Solutions October 20, 2005

3.84/ Hypergeometric with r = 5, N = 10, n = 3 (think green and “not green”), so

P (Y = 3) = \binom{5}{3}\binom{5}{0} \binom{10}{3} = \frac{1}{12}.

3.88/ Y is the number of nonlocal firms. a) P (Y = 1) = \binom{2}{1}\binom{4}{2} \binom{6}{3} = \frac{3}{5}. b) P (Y ≥ 1) =

\frac{3}{5} + P (Y = 2) = \frac{3}{5} + \binom{2}{2}\binom{4}{1} \binom{6}{3} = \frac{4}{5}. c) P (Y ≤ 1) = P (Y = 0) + P (Y = 1) = \frac{4}{5}.

3.89/ For a) hypergeometric with N = 10, n = 3, r = 2 so table of values of the pdf is

\begin{matrix}y & 0 & 1 & 2 \\ P (Y = y) & 14/30 & 14/30 & 2/30 \end{matrix}

For b), N = 10, n = 3, r = 4, so

\begin{matrix}y & 0 & 1 & 2 & 3 \\ P (Y = y) & 5/30 & 15/30 & 9/30 & 1/30 \end{matrix}

3.97/ Y is Poisson with λ = 2.

a) P (Y = 4) = \frac{2^{4}}{4!} e^{−2} ≐ .090.

b) Using the table of Poisson probabilities in text: P (Y ≥ 4) = 1−P (Y ≤ 3) = 1−.857 = .143.

c) P (Y < 4) = P (Y ≤ 3) = .857.

d) P (Y ≥ 4|Y ≥ 2) = .143/.594 = .241 (using definition of conditional probability).

3.101/ Poisson with λ = 4. P (Y ≤ 4) = .629 from the text’s table of Poisson probabilities.

3.102/ Method 1 – “direct approach”: Let I be the number of cars arriving at entrance I and II similarly for entrance II. Let T = I + II be the total number. We can get T = 3 in four different ways (I = 3 and II = 0, or I = 2 and II = 1, etc.) Then using independence:

P (T = 3) = ∑_{k=0}^{3} P (I = k)P (II = 3 − k)

= ∑_{k=0}^{3} \frac{3^{k}e^{−3}}{k!} \frac{4^{3−k}e^{−4}}{(3 − k)!}

= e^{−7} (\frac{9}{2} + 18 + 24 + \frac{32}{3})

≐ .0521 Method 2 – “clever”: The fact that there are two entrances is really irrelevant for the total number T of cars entering if the numbers from the two entrances are independent. It should be the same as a situation where you have a Poisson distribution with λ = 3+4 = 7 cars entering per hour. Then P (T = 3) = \frac{7^{3}e^{−7}}{3!} ≐ .0521. (Same as above!) Comment: In fact, this problem illustrates a general fact about Poisson random variables. If you have X Poisson with parameter λ and Y Poisson with parameter μ and X, Y are independent, then Z = X + Y also has a Poisson distribution with parameter λ + μ. One proof comes from an argument like the “direct approach” above:

P (Z = z) = ∑_{x=0}^{z} P (X = x)P (Y = z − x)

= ∑_{x=0}^{z} P (X = x)P (Y = z − x)

= ∑_{x=0}^{z} \frac{λ^{x}e^{−λ}}{x!} \frac{λ^{z−x}e^{−λ}}{(z − x)!}

= e^{−(λ+μ)} ∑_{x=0}^{z} \frac{1}{x!(z − x)!}λ^{x}μ^{z−x}

= \frac{e^{−(λ+μ)}}{z!} ∑_{x=0}^{z} \binom{z}{x}λ^{x}μ^{z−x}

= \frac{e^{−(λ+μ)}}{z!} (λ + μ)^{z}

(using the binomial theorem). This is exactly the right probability mass function for a Poisson random variable with parameter λ + μ. There’s another proof via the moment generating functions too(!) If X, Y are independent, and Z = X + Y, then

m_{Z}(t) = E(e^{tX+tY}) = E(e^{tX})E(e^{tY}) = e^{λ(e^{t}−1)}e^{μ(e^{t}−1)} = e^{(λ+μ)(e^{t}−1)}

By the uniqueness theorem, Z has a Poisson distribution with parameter λ + μ.

3.114/

a) We have

p(y)/p(y − 1) = \frac{\frac{λ^{y}}{y!} e^{−λ}}{\frac{λ^{y−1}}{(y−1)!}e^{−λ}} = \frac{λ}{y}

b) Hence p is increasing for y < λ, and decreasing for y > λ.

c) p(y) is maximized with y = ⌊λ⌋ (the greatest integer less than or equal to λ).

3.115/ By the binomial theorem,

m(t) = E(e^{tY}) = ∑_{y=0}^{n} e^{ty}p^{y}q^{n−y}

= ∑_{y=0}^{n}(e^{t}p)^{y}q^{n−y}

= (pe^{t} + q)^{n}.

3.116/ The idea of this problem is to use the moment generating function for a binomial Y to give a different derivation of the formulas we know for E(Y) and V (Y). By the chain rule for derivatives, m′(t) = n(pe^{t} + q)^{n−1}pe^{t}. So E(Y) = m′(0) = np. Then

m′′(t) = n(n − 1)(pe^{t} + q)^{n−2}p^{2}e^{2t} + n(pe^{t} + q)^{n−1}pe^{t}

so E(Y^{2}) = m′′(0) = n(n − 1)p^{2} + np. Then

V (Y) = E(Y^{2}) − E(Y)^{2} = n(n − 1)p^{2} + np − n^{2}p^{2} = −np^{2} + np = np(1 − p) = npq.

3.121/ By the Uniqueness Theorem for the moment generating function, and the formulas from class and the text for moment generating functions for standard types of random variables,

a) Binomial, n = 5, p = 1/3.

b) Geometric, p = 1/2.

c) Poisson, λ = 2.

3.123/ a) E(Y) = m′(0) = 1/6+2/3+3/2 = 7/3. b) E(Y^{2}) = m′′(0) = 1/6+4/3+9/2 = 6. Then V (Y) = E(Y^{2})−E(Y)^{2} = 6−(7/3)^{2} = 5/9 c) By thinkning about the form of m_{Y} (t), we see that Y is a discrete random variable with P (Y = 1) = 1/6, P (Y = 2) = 2/6, and P (Y = 3) = 3/6 (all other probability values zero).

3.124/ Since e^{bt} is not random (it’s a fixed function of t),

m_{aY +b}(t) = E(e^{(aY +b)t}) = E(e^{bt}e^{(at)Y}) = e^{bt}E(e^{(at)Y}) = e^{bt}m_{Y} (at).