Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 7 – November 5, 2003

Text Problems:

4.69/ a) The probability that the magnitude Y of a single earthquake will exceed 3 is

P (Y > 3) = ∫_{3}^{∞} \frac{1}{2.4}e^{−y/2.4} dy = −e^{−y/2.4}|_{3}^{∞} = 0 + e^{−3/2.4} ≐ .287

b) The probability that the magnitude Y of an earthquake will fall between 2 and 3 is

P (2 < Y < 3) = ∫_{2}^{3} \frac{1}{2.4}e^{−y/2.4} dy = −e^{−y/2.4}|_{2}^{3} = −e^{−3/2.4} + e^{−2/2.4} ≐ .148

4.70/ Similarly to 4.69 a, the probability that a single earthquake has magnitude greater than 5 is

∫_{5}^{∞} \frac{1}{2.4}e^{−y/2.4} dy = −e^{−y/2.4}|_{5}^{∞} = e^{−5/2.4} ≐ .1245

Now, assuming the next 10 earthquakes are independent events, we can model the number exceeding 5 on the Richter scale as a binomial (discrete) random variable with n = 10, p = .1245. The probability that this number is at least one is

∑_{y=1}^{10} \binom{10}{y}(.1245)^{y}(.8755)^{10−y} = 1 − (.8755)^{10} ≐ .7354.

(Better get your home insurance in order!!)

4.71/ a) This part is similar to 4.69 a: Let Y be the demand

P (Y > 200) = ∫_{200}^{∞} \frac{1}{100}e^{−y/100} dy = −e^{−y/100}|_{200}^{∞} = 0 + e^{−200/100} ≐ .1353

b) We want to find the value c so that P (Y > c) = .01. This gives the equation

.01 = ∫_{c}^{∞} \frac{1}{100}e^{−y/100} dy = e^{−c/100}

Taking natural logarithms of both sides we get

c = −100 ln(.01) ≐ 460.5

4.72/ Since Y is exponential with mean β = 10, we know from class or Theorem 4.10 in the book that E(Y) = β = 10 and E(Y^{2}) = β^{2} + β^{2} = 200. Therefore

E(C) = E(100 + 40Y + 3Y^{2}) = 100 + 40E(Y) + 3E(Y^{2}) = 100 + 400 + 600 = 1100 To find the variance of the cost, we will use V (C) = E(C^{2}) − (E(C))^{2}. To find E(C^{2}), multiply out first:

C^{2} = (100 + 40Y + 3Y^{2})^{2} = 10000 + 800Y + 2200Y^{2} + 240Y^{3} + 9Y^{4}

We know E(Y) and E(Y^{2}) from the above. To find E(Y^{3}) andE(Y^{4}), we will use the substitution u = y/10 on the corresponding integrals to recognize them as constants times Γ-function values:

\begin{aligned}E(Y^{3}) & = ∫_{0}^{∞} y^{3} \frac{1}{10}e^{−y/10} dy = 10^{3} ∫_{0}^{∞} u^{3}e^{−u} du = 10^{3}Γ(4) \\ E(Y^{4}) & = ∫_{0}^{∞} y^{4} \frac{1}{10}e^{−y/10} dy = 10^{4} ∫_{0}^{∞} u^{4}e^{−u} du = 10^{4}Γ(5) \end{aligned}

Hence

E(C^{2}) = 10000 + 800 · 10 + 2200 · 200 + 240 · 6000 + 9 · 240000 = 4, 130, 000

Then

V (C) = 4, 130, 000 − (1100)^{2} = 2, 920, 000.

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 distribution function for a geometric random variable with p = 1 − e^{\frac{−1}{β}}.

4.83/ Using Theorem 4.8, μ = αβ = 3.2 and σ^{2} = αβ^{2} = 6.4.

4.87 The idea for this one is similar to what we did above in 4.72. 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.

4.91/ From the form of f(y), we recognize that Y has a Beta-distribution with parameters α = 4 and β = 3. Hence

k = \frac{1}{B(4, 3)} = \frac{Γ(7)}{Γ(4)Γ(3)} = \frac{6!}{3!2!} = 60.

4.97/ The idea is similar to 4.72 and 4.87. Here the moments E(Y^{k}) are ratios of Betafunction values. We find using Theorem 4.11 and similar calculations:

\begin{aligned}E(Y) & = \frac{B(2, 2)}{B(1, 2)} = 1/3 \\ E(Y^{2}) & = \frac{B(3, 2)}{B(1, 2)} = 1/6 \\ E(Y^{3}) & = \frac{B(4, 2)}{B(1, 2)} = 1/10 \\ E(Y^{4}) & = \frac{B(5, 2)}{B(1, 2)} = 1/15 \end{aligned}

Hence for C = 10 + 20Y + 4Y^{2},

E(C) = 10 + 20 · (1/3) + 4 · (1/6) = 52/3

Then V (C) = V (20Y + 4Y^{2}), so

E(20Y + 4Y^{2}) = 20 · (1/3) + 4 · 1/6 = 22/3,

E((20Y + 4Y^{2})^{2}) = 400 · (1/6) + 160 · (1/10) + 16 · (1/15) = 1256/15

and

V (C) = 1256/15 − (22/3)^{2} = 1348/45 ≐ 29.96

4.98/ For a Beta-distributed Y with parameters α and β, as in 4.97, we have

E(Y^{2}) = \frac{1}{B(α, β)} ∫_{0}^{1} y^{2} · y^{α−1}(1 − y)^{β−1} dy

= \frac{B(α + 2, β)}{B(α, β)}

= \frac{Γ(α + 2)Γ(β)Γ(α + β)}{Γ(α + β + 2)Γ(α)Γ(β)}

= \frac{(α + 1)αΓ(α)Γ(β)Γ(α + β)}{(α + β + 1)(α + β)Γ(α + β)Γ(α)Γ(β)}

= \frac{(α + 1)α}{(α + β + 1)(α + β)}

We already know E(Y) = \frac{α}{α+β}. Hence

V (Y) = \frac{(α + 1)α}{(α + β + 1)(α + β)} − (\frac{α}{α + β} )^{2}

= \frac{(α + β)(α + 1)α − α^{2}(α + β + 1)}{(α + β + 1)(α + β)^{2}}

= \frac{αβ}{(α + β)^{2}(α + β + 1)}

which is what we wanted to show. 4.101/From the form of f(y) we recognize that Y has a Beta-distribution with α = 3, β = 5. Hence a) c = \frac{1}{B(3,5)} = 105, and b) E(Y) = \frac{3}{3+5} = 3/8, using Theorem 4.11. Additional Problems 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

B) The idea of this proof is similar to what we did for A). Now we will use induction on β, with α a fixed, but arbitrary, positive integer. In other words we must first show the base case that

(3)  \frac{1}{B(α, 1)} ∫_{0}^{p} y^{α−1} dy = ∑_{n=α}^{α} \binom{α}{n}p^{n}(1 − p)^{α−n}.

The integral on the left of (3) is an easy one, we just get:

\frac{1}{B(α, 1)α}y^{α}|_{0}^{p} = \frac{Γ(α + 1)}{αΓ(α)}p^{α} = p^{α}

(since Γ(α + 1) = αΓ(α)). The sum on the right of (3) is again just one term: 1 · p^{α}. So our base case is established.

Now, for the induction step we assume that the formula is true for β = k (and α = anything):

(4)  \frac{1}{B(α, k)} ∫_{0}^{p} y^{α−1}(1 − y)^{k−1} dy = ∑_{n=α}^{α+k−1} \binom{α + k − 1}{n}p^{n}(1 − p)^{α+k−1−n}.

Start from the integral with β = k + 1:

\frac{1}{B(α, k + 1)} ∫_{0}^{p} y^{α−1}(1 − y)^{k} dy

Integrate by parts with u = (1 − y)^{k} (idea is to lower the exponent here to match the integral on the left of (4). This makes dv = y^{α−1} dy. Then du = −k(1 − y)^{k−1}, and v = y^{α}/α. Applying the parts formula and the induction hypothesis (with α replaced by α + 1), we get

= \frac{1}{B(α, k + 1)} (\frac{y^{α}(1 − y)^{k}}{α} |_{0}^{p} + \frac{k}{α} ∫_{0}^{p} y^{α}(1 − y)^{k−1} dy)

= \binom{α + k}{α}p^{α}(1 − p)^{k} + ∑_{n=α+1}^{α+1+k−1} \binom{α + 1 + k − 1}{n}p^{n}(1 − p)^{α+1+k−1−n}

= \binom{α + k}{α}p^{α}(1 − p)^{k} + ∑_{n=α+1}^{α+k} \binom{α + k}{n}p^{n}(1 − p)^{α+k−n}

= ∑_{n=α}^{α+k} \binom{α + k}{n}p^{n}(1 − p)^{α+k−n}

which is what we want.

2) Using the binomial table, with p = .3, α = 4, β = 7,

P (0 ≤ Y ≤ .3) = ∑_{n=4}^{10} \binom{10}{n}(.3)^{n}(.7)^{10−n} = 1 − .650 = .350.