Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 8 November 5, 2009

4.129. Instead of computing the expected values directly from the integral formulas, we can use the general results for β-distributed random variables from Theorem 4.11. We have

E(C) = E(10 + 20Y + 4Y^{2}) = 10 + 20E(Y) + 4E(Y^{2}).

From Theorem 4.11, we have E(Y) = \frac{α}{α+β} = \frac{1}{3}. Then

E(Y^{2}) = V (Y) + E(Y)^{2} = \frac{αβ}{(α + β)^{2}(α + β + 1)} + \frac{1}{9} = \frac{1}{18} + \frac{1}{9} = \frac{1}{6}.

So E(C) = 10 + \frac{20}{3} + \frac{4}{6} = \frac{52}{3}. To get V (C) = E(C^{2}) − E(C)^{2}, we can proceed as follows: E(C^{2}) = 100 + 400E(Y) + 480E(Y^{2}) + 160E(Y^{3}) + 16E(Y^{4}). We already know E(Y), E(Y^{2}) from the above. To get E(Y^{3}) and E(Y^{4}) proceed like this:

E(Y^{3}) = 2 ∫_{0}^{1} y^{3}(1 − y) dy.

Apart from the constant multiple, this is the integral of a beta density function with α = 4 and β = 2. So

E(Y^{3}) = 2B(4, 2) = 2\frac{Γ(4)Γ(2)}{Γ(6)} = 2\frac{3!1!}{5!} = \frac{1}{10}.

Similarly,

E(Y^{4}) = 2B(5, 2) = 2\frac{Γ(5)Γ(2)}{Γ(7)} = 2\frac{4!1!}{6!} = \frac{1}{15}.

So then

V (C) = 10 + \frac{400}{3} + \frac{480}{6} + \frac{160}{10} + \frac{16}{15} − (\frac{52}{3})^{2} = \frac{1348}{45}.

4.130. We use the same “trick” as in the last problem. We know E(Y) = \frac{α}{α+β} from Theorem 4.11 and work in class.

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.

4.133. (a) This is the form for a beta density function with α = 3 and β = 5, so c = \frac{1}{B(3,5)} = \frac{7!}{2!4!} = 105.

(b) E(Y) = \frac{3}{8} = .375 from Theorem 4.11.

(c) The standard deviation is

σ = \sqrt{\frac{15}{64 · 9}} ≐ .1614.

(d) We have:

P (Y > .375 + 2(.1614)) = ∫_{.6978}^{1} 105y^{2}(1 − y)^{4} dy ≐ .02972.

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) is the moment-generating function for a gamma-distributed random variable with α = 2, β = 4.

(b) is the moment-generating function for a gamma-distributed random variable with α = 1, β = 3.2. Recall that the gamma-distributed variables with α = 1 are also called exponential. (c) is normal with μ = −5 and σ^{2} = 12.

4.141. The moment generating function is

m_{Y} (t) = E(e^{ty})

= ∫_{θ_{1}}^{θ_{2}} \frac{e^{ty}}{θ_{2} − θ_{1}} dy

= \frac{e^{tθ_{2}} − e^{tθ_{1}}}{t(θ_{2} − θ_{1})} .

4.147. By Tchebysheff’s Theorem, P (|Y − μ| ≤ kσ) ≥ 1 − 1/k^{2}. To get 1 − 1/k^{2} = .75, we take k = 2. 2σ = 1oz implies σ = .5oz.

4.149. We can shift and scale to consider a uniform random variable on [0, 1], where we know μ = \frac{1}{2} and σ^{2} = \frac{1}{12}. Hence μ ± 2σ = \frac{1}{2} ± 1\sqrt{3}. Since \frac{1}{\sqrt{3}} > \frac{1}{2}, μ + 2σ > 1 and μ − 2σ < 0. Hence P (|Y − μ| ≤ 2σ) = 1. This compares with P (|Y − μ| ≤ 2σ) ≥ \frac{3}{4} by Tchebysheff. It is greater than the P (|Y − μ| ≤ 2σ) ≐ .95 according to the empirical rule.

4.150. For an exponential random variable, μ = β and σ = β, so P (|Y − β| ≤ 2β) = P (−β < Y < 3β) = P (0 < Y < 3β). This is

∫_{0}^{3β} \frac{1}{β} e^{\frac{−y}{β}} dy = 1 − e^{−3}
≐ .9502.

This compares with P (|Y − μ| ≤ 2σ) ≥ \frac{3}{4} by Tchebysheff, and is almost exactly the same as the P (|Y − μ| ≤ 2σ) ≐ .95 predicted by the empirical rule.

5.4. (a) is clear.

(b) F (1, 2) = .05. This is the probability that the child did not survive if a car-set was used.

5.6. (a) The region described by the inequality Y_{1}−Y_{2} > .5 is the region in the unit square below the line Y_{2} = Y_{1} − .5, which intersects the edges at (Y_{1}, Y_{2}) = (.5, 0) and (1, .5). The given probability is

∫_{.5}^{1} ∫_{0}^{y_{1}−.5} dy_{2} dy_{1} = \frac{(y_{1} − .5)^{2}}{2} |_{.5}^{1} = .125.

(b) The region is now the part of the unit square underneath the hyperbola Y_{1}Y_{2} = .5. Note that to set up the double integrals, we must split the region of integration in this case.

P (Y_{1}Y_{2} < .5) = ∫_{0}^{.5} ∫_{0}^{1} dy_{2} dy_{1} + ∫_{.5}^{1} ∫_{0}^{.5/y_{1}} dy_{2} dy_{1}
= .5 + .5 ln(y_{1})|_{.5}^{1}
= .5(1 − ln(.5))
≐ .8465.

5.7. The density is only nonzero on the first quadrant of the (y_{1}, y_{2})-plane. (a) The inequalities Y_{1} < 1, Y_{2} > 5 describe an infinite vertical strip.

P (Y_{1} < 1, Y_{2} > 5) = ∫_{0}^{1} ∫_{5}^{∞} e^{−y_{1}−y_{2}} dy_{2} dy_{1}
= ∫_{0}^{1} −e^{−y_{1}−y_{2}} |_{y_{2}=5}^{y_{2}=∞} dy_{1}
= ∫_{0}^{1} e^{−y_{1}−5} dy_{1}
= −e^{−y_{1}−5}|_{0}^{1}
= e^{−5} − e^{−6}
≐ .004259.

(b) The inequality Y_{1} + Y_{2} < 3 describes the region in the the first quadrant below the line Y_{2} = −Y_{1} + 3:

P (Y_{1} + Y_{2} < 3) = ∫_{0}^{3} ∫_{0}^{3−y_{1}} e^{−y_{1}−y_{2}} dy_{2} dy_{1}
= ∫_{0}^{3} e^{−y_{1}} − e^{−3} dy_{1}
= 1 − 4e^{−3}
≐ .8009.

5.8. (a) We must have 1 = ∫_{0}^{1} ∫_{0}^{1} ky_{1}y_{2} dy_{2} dy_{1} = \frac{k}{4}. So k = 4.

(b) b) The joint cdf is

F (y_{1}, y_{2}) = P (Y_{1} ≤ y_{1}, Y_{2} ≤ y_{2}) = \begin{cases}y_{1}^{2}y_{2}^{2} & if 0 ≤ y_{1}, y_{2} ≤ 1 \\ y_{2}^{2} & if y_{1} ≥ 1, 0 ≤ y_{2} ≤ 1 \\ y_{1}^{2} & if y_{2} ≥ 1, 0 ≤ y_{1} ≤ 1 \\ 1 & if y_{1}, y_{2} ≥ 1 \\ 0 & otherwise. \end{cases}

c) P (Y_{1} ≤ 1/2, Y_{2} ≤ 3/4) = ∫_{0}^{1/2} ∫_{0}^{3/4} 4y_{1}y_{2} dy_{2} dy_{1} = (1/2)^{2}(3/4)^{2} = 9/64.