Mathematics 375 – Probability Theory Solutions for Problem Set 7 November 4, 2011

4.88. (a) Let Y be the magnitude of the earthquake. The density for Y is then f(y) = \frac{e^{−y/2.4}}{2.4} for y > 0.

P (Y > 3) = ∫_{3}^{∞} \frac{e^{−y/2.4}}{2.4} dy = e^{−3/2.4} ≐ .2865.

(b) Similarly.

P (2 < Y < 3) = ∫_{2}^{3} \frac{e^{−y/2.4}}{2.4} dy = e^{−2/2.4} − e^{−3/2.4} ≐ .1481.

4.95. (a) We have by the definition of the discrete random variable X:

P (X = k) = ∫_{k−1}^{k} \frac{1}{β} e^{\frac{−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}} ) .

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

4.96. (a) We want 1 = k ∫_{0}^{∞} y^{3}e^{−y/2} dy. The best way to do this part is to note that this looks like the Gamma density in the case α = 4 and β = 2. Hence the constant k must be:

k = \frac{1}{2^{4}Γ(4)} = \frac{1}{16 · 3!} = \frac{1}{96}.

(Harder way: Integrating by parts three times, or using a table of integrals, we find k = 96.) (b) This is a χ^{2} random variable with ν = 8. (c) The mean and standard deviation follow by the formulas from Theorem 4.9 – μ = 8, σ^{2} = 16. (d) Not assigned, but here is how you could do this. The interval μ ± 2σ is [0, 16]. We have

∫_{0}^{16} \frac{y^{3}e^{−y/2}}{96} dy ≐ .9576

(I used Maple to compute this approximate value; you could also integrate by parts three times).

4.105. (a) If Y is the four-week total, then by Theorem 4.8, E(Y) = 3.2 and V (Y) = 6.4. (b) Not assigned, but using numerical integration, this is how it would go: This probability is the value

P (Y > 4) = ∫_{4}^{∞} \frac{y^{2.2}e^{−y/2}}{Γ(3.2)2^{3.2}} dy ≐ .2896.

4.111. We have, letting u = y/β:

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

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

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

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

= \frac{β^{a}Γ(a + α)}{Γ(α)},

since the integral on the right on the second to last line computes Γ(a + α). (b) We must have a + α > 0, since we only defined Γ(x) for x > 0. (For those of you taking complex analysis, there is a way to define Γ(x) for some negative real x as well, using the technique of analytic continuation, but we are not doing that in this course.) (c) If we take a = 1, then by the functional equation for the Γ function,

\frac{βΓ(1 + α)}{Γ(α)} = \frac{βαΓ(α)}{Γ(α)} = αβ.

(d) With a = 1/2, we only need α > 0. We get

E(\sqrt{Y}) = \frac{\sqrt{β}Γ(α + 1/2)}{Γ(α)}.

(e) The cases where a is an integer can be simplified using the functional equation for the Γ function. With a = −1, we need α > 1, and then

E(1/Y) = \frac{Γ(α − 1)}{βΓ(α)} = \frac{Γ(α − 1)}{β(α − 1)Γ(α − 1)} = \frac{1}{β(α − 1)}.

With a = −1/2, we need α > 1/2 and then

E(1/\sqrt{Y}) = \frac{Γ(α − 1/2)}{\sqrt{β}Γ(α)}.

Finally, with a = −2, we need α > 2 and then

E(1/Y^{2}) = \frac{Γ(α − 2)}{β^{2}Γ(α)} = \frac{Γ(α − 2)}{β^{2}(α − 1)(α − 2)Γ(α − 2)} = \frac{1}{β^{2}(α − 1)(α − 2)}.

4.128. We recognize the density as the Beta density with α = 1 and β = 3. We want to set the budget for repairs at a level b so that P (Y > b) = ∫_{b}^{1} 3(1 − y)^{2} dy = .1. This gives the equation

.1 = −(1 − y)^{3}|_{b}^{1} = (1 − b)^{3}

Hence we want b = .5358 × $100 = $53.58.

4.129. The mean of C can be computed from the formulas for E(Y) and V (Y) = E(Y^{2}) + (E(Y))^{2} from Theorem 4.11:

E(C) = E(10 + 20Y + 4Y^{2})
= 10 + 20E(Y) + 4E(Y^{2})
= 10 + 20(1/3) + 4(1/18 + (1/3)^{2})
= 52/3.

So the average cost is about ≐ 17.3 × $100 = $1730. For the variance of the cost we can compute like this:

V (C) = E(C^{2}) − (E(C))^{2}

= ∫_{0}^{1}(10 + 20y + 4y^{2})^{2} · 2(1 − y) dy − (52/3)^{2}

= \frac{1652}{5} − (\frac{52}{3})^{2}

≐ 29.96

This would correspond to V (C) = 29.96 × 100^{2} ≐ 299600 dollars squared.

4.130. We know E(Y) = \frac{α}{α+β} from Theorem 4.11 and work in class. We compute similarly,

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

= \frac{1}{B(α, β)} ∫ 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)Γ(β)Γ(α + β + 2)}.

Then using the functional equation for the Γ-function

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

= \frac{(α + 1)α}{(α + β + 1)(α + β)}.

So then

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

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

= \frac{αβ}{(α + β)^{2}(α + β + 1)},

which is what we wanted to show.

4.136. (a) Y has an exponential distribution with β = θ. So by Example 4.13 with α = 1 and β = θ, the moment generating function is m_{Y} (t) = \frac{1}{1−θt}.

(b) Then

E(Y) = m′_{Y} (0) = \frac{θ}{(1 − θt)^{2}} |_{t=0} = θ

and

E(Y) = m′′_{Y} (0) = \frac{2θ^{2}}{(1 − θt)^{3}} |_{t=0} = 2θ^{2}

Hence V (Y) = E(Y^{2}) − E(Y)^{2} = 2θ^{2} − θ^{2} = θ^{2}.

4.137. Letting U = aY + b, since U is a function of Y, we find its moment-generating function by integrating e^{t(ay+b)} times the density function f(y):

m_{U}(t) = ∫_{−∞}^{∞} e^{t(ay+b)}f(y) dy
= ∫_{−∞}^{∞} e^{(at)y} · e^{tb}f(y) dy
= e^{tb} ∫_{−∞}^{∞} e^{(at)y}f(y) dy
= e^{tb}m_{Y} (at)

Then to find E(U) and V (U), we compute as usual, using the product and chain rules (note: \frac{d}{dt}m_{Y} (at) = m′_{Y} (at) · a) to find the necessary derivatives:

E(U) = m′_{U}(0)

= \frac{d}{dt}(e^{tb}m_{Y} (at))|_{t=0}

= (be^{tb}m_{Y} (at) + e^{tb}m′_{Y} (at)a)|_{t=0}

= bm_{Y} (0) + am′_{Y} (0)

= b + aμ, since m_{Y} (0) = 1 and m′_{Y} (0) = E(Y) = μ. Then

E(U^{2}) = m′′_{U}(0)

= \frac{d}{dt}(be^{tb}m_{Y} (at) + e^{tb}m′_{Y} (at)a)|_{t=0}

= (b^{2}e^{tb}m_{Y} (at) + 2be^{tb}m′_{Y} (at)a + e^{t}bm′′_{Y} (t)a^{2})|_{t=0}

= b^{2} + 2abμ + a^{2}E(Y^{2}).

Hence

V (U) = b^{2} + 2abμ + a^{2}E(Y^{2}) − (b + aμ)^{2} = a^{2}(E(Y^{2}) − μ^{2}) = a^{2}σ^{2}.

4.139. Using the formula of Exercise 4.138 and the moment-generating function for a normal random variable, we have

m_{X}(t) = e^{4t}m_{Y} (−3t) = e^{4t} · e^{\frac{σ^{2}(−3t)^{2}}{2} +μ(−3t)} = e^{\frac{(3σ)^{2}t^{2}}{2} +(4−3μ)t}

This is the moment-generating function for a normal random variable with mean 4−3μ and variance 9σ^{2}. Hence by the uniqueness theorem we quoted in class, X is normally distributed.

4.140. Using the standard forms (for instance from the table inside the back cover of the textbook),

(a) Gamma(2, 4) (b) Gamma(1, 3.2) = Exponential(3.2) (c) Normal(−5, 12).

4.145. Note that this is sort of a “reverse” or “negative” exponential density. (a) We have

E(e^{3Y/2}) = ∫_{−∞}^{0} e^{3y/2} · e^{y} dy

= ∫_{−∞}^{0} e^{5y/2} dy

= \frac{2}{5}e^{5y/2}|_{−∞}^{0}

= \frac{2}{5}.

(b) The moment generating function is found by a similar computation:

E(e^{tY}) = ∫_{−∞}^{0} e^{ty} · e^{y} dy

= ∫_{−∞}^{0} e^{(1+t)y} dy

= \frac{1}{1 + t}e^{(1+t)y}|_{−∞}^{0}

= \frac{1}{1 + t}.

Note that 1 + t > 0 in order for the integral to converge. So the mgf is only defined for t > −1. (c) V (Y) = m′′_{Y} (0) − (m′_{Y} (0))^{2}. Since

m′_{Y} (t) = −(1 + t)^{−2}
m′′_{Y} (t) = 2(1 + t)^{−3},

we have m′_{Y} (0) = −1, and m′′_{Y} (0) = 2, so V (Y) = 2 − (−1)^{2} = 1.

4.146. If k = \sqrt{10} ≐ 3.162, then Tchebysheff’s theorem says

P (|Y − μ| ≥ 3.162σ) ≤ \frac{1}{10}.

So one possible interval is μ ± 3.162σ, or 25000 ± 12648, or [12352, 37648].

4.147. By Tchebysheff’s theorem, P (|Y − μ| < 2σ) ≥ \frac{3}{4}. So if σ = 1/2, the interval μ ± 2σ will be μ ± 1.