Mathematics 375 – Probability and Statistics I Solutions for Problem Set 7 October 28, 2005

Text Problems:

4.73/ From the given information, then pdf for Y is f(y) = \{ \begin{matrix}\frac{1}{44}e^{−y/44} & if y > 0 \\ 0 & otherwise \end{matrix}. a)

P (Y < 31) = ∫_{0}^{31} \frac{1}{44}e^{−y/44} dy = 1 − e^{−31/44} ≐ .506. b) From the general formula for exponentially-distributed Y, V (Y) = β^{2} = 44^{2} = 1936.

4.75/ We have by the definition of the discrete random variable X:

P (X = k) = ∫_{k−1}^{k} \frac{1}{β} e^{−y/β} dy
= −e^{\frac{−y}{β}} |_{k−1}^{k}
= −e^{\frac{−k}{β}} + e^{\frac{−(k−1)}{β}}
= (e^{\frac{−(k−1)}{β}})(1 − e^{\frac{−1}{β}})
= (e^{\frac{−1}{β}})^{k−1}(1 − e^{\frac{−1}{β}})

This is the form (q^{k−1}p) for the probability function for a geometric random variable with p = 1 − e^{\frac{−1}{β}}.

4.77/ Exponential distribution with mean β = 4, so ∫_{4}^{∞} e^{−y/4}/4dy = e^{−1} ≐ .3679.

4.78/ We want to determine y_{0} so that P (Y > y_{0}) = .05. This says

.05 = P (Y > y_{0}) = ∫_{y_{0}}^{∞} \frac{1}{4}e^{−y/4} dy = e^{−y_{0}/4}.

We solve .05 = e^{−y_{0}/4} for y_{0} by taking natural log of both sides: y_{0} = −4 ln(.05) ≐ 11.98.

4.86/ a) Using the general formulas for a Γ-distribution, μ = E(Y) = αβ = 1000 · 20 = 20000 (units $ ) and V (Y) = αβ^{2} = 400000 (units $^{2}). b) From part a, σ = \sqrt{V (Y)} ≐ 632.5. This means that 40000 ≐ μ + 31σ so it is very unlikely that any income is that large. (Another way to answer this would be to compute the integral ∫_{40000}^{∞} f(y) dy for the gamma pdf. This would come out extremely small!)

4.87/ First, let’s derive the general formula for E(Y^{k}) when Y has a Gamma-distribution with parameters α, β. We apply the substitution u = \frac{y}{β}, so dy = β du, then apply the functional equation Γ(x + 1) = xΓ(x) repeatedly:

E(Y^{k}) = \frac{1}{Γ(α)β^{α}} ∫_{0}^{∞} y^{k} · y^{α−1}e^{\frac{−y}{β}} dy

= \frac{1}{Γ(α)β^{α}} ∫_{0}^{∞} y^{α+k−1}e^{\frac{−y}{β}} dy

= \frac{1}{Γ(α)β^{α}} · β^{α+k} ∫_{0}^{∞} u^{α+k−1}e^{−u} du

= \frac{1}{Γ(α)β^{α}} · β^{α+k} · Γ(α + k)

= β^{k}(α + k − 1)(α + k − 2) ⋯ (α + 1)(α)

In our problem, with α = 3, β = 2, we get

\begin{aligned}E(Y) & = αβ = 6 \\ E(Y^{2}) & = (α + 1)(α)β^{2} = 48 \\ E(Y^{3}) & = (α + 2)(α + 1)αβ^{3} = 480 \\ E(Y^{4}) & = (α + 3)(α + 2)(α + 1)αβ^{4} = 5760 \end{aligned}

Then

E(L) = 30E(Y) + 2E(Y^{2}) = 30 · 6 + 2 · 48 = 276

and

E(L^{2}) = 900E(Y^{2}) + 120E(Y^{3}) + 4E(Y^{4}) = 900 · 48 + 120 · 480 + 4 · 5760 = 123840.

Hence

V (C) = E(L^{2}) − (E(L))^{2} = 123840 − 276^{2} = 47664.

4.92/ P (Y > .40) = ∫_{.40}^{1} 12y^{2}(1 − y) dy = (4y^{3} − 3y^{4}|_{.40}^{1} ≐ .8208.

4.93/ Being “creatively lazy,” we do not need to grind through the “from-scratch” computation of the expected value and variance because we recognize that Y has a Beta distribution with α = 3 and β = 2 from the formula for the pdf. This means we can use the general formulas to obtain E(Y) = α/(α + β) = 3/5 and V (Y) = αβ/((α + β)^{2}(α + β + 1)) = 6/(25 · 6) = 1/25.

Additional Problem

A) 1) We must show that when α is a positive integer,

(1)  \frac{1}{Γ(α)} ∫_{λ}^{∞} y^{α−1}e^{−y} dy = ∑_{n=0}^{α−1} \frac{λ^{n}}{n!} e^{−λ}.

We will use induction on α. The base case is α = 1. Then the left of (1) is

1 · ∫_{λ}^{∞} e^{−y} dy = e^{−λ}.

With α = 1, the sum on the right of (1) is:

∑_{n=0}^{0} \frac{λ^{n}}{n!} e^{−λ} = e^{−λ}

So the base case is established.

Now for the induction step, assume that (1) is true for α = k. In other words, our induction hypothesis is:

(2)  \frac{1}{Γ(k)} ∫_{λ}^{∞} y^{k−1}e^{−y} dy = ∑_{n=0}^{k−1} \frac{λ^{n}}{n!} e^{−λ}.

Consider the integral on the left of (1) for α = k + 1:

\frac{1}{Γ(k + 1)} ∫_{λ}^{∞} y^{k}e^{−y} dy

Apply integration by parts with u = y^{k}, and dv = e^{−y} dy. We get:

\begin{aligned}\frac{1}{Γ(k + 1)} ∫_{λ}^{∞} y^{k}e^{−y} dy & = \frac{1}{Γ(k + 1)} (−y^{k}e^{−y}|_{λ}^{∞} + k ∫_{λ}^{∞} y^{k−1}e^{−y} dy) \\ & = \frac{λ^{k}}{k!} e^{−λ} + \frac{1}{Γ(k)} ∫_{λ}^{∞} y^{k−1}e^{−y} dy \end{aligned}

Now we use the induction hypothesis (2) on the remaining integral. We get that this is equal to

\frac{λ^{k}}{k!} e^{−λ} + ∑_{n=0}^{k−1} \frac{λ^{n}}{n!} e^{−λ} = ∑_{n=0}^{k} \frac{λ^{n}}{n!} e^{−λ}

Hence our formula also works with α = k + 1, and by induction we conclude that it works for all positive integers α.

2) Using the Poisson table in the text with λ = 5, α = 3

∑_{n=0}^{2} \frac{5^{n}}{n!} e^{−5} = .125