Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 8 November 16, 2009

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}.

4.139. 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 with mean μ_{X} = 4 − 3μ_{Y} and variance σ_{X}^{2} = 9σ_{Y}^{2}

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.

4.147. 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) = 1. This is sum of all of the entries in the table, since F (1, 2) = P (Y_{1} ≤ 1, Y_{2} ≤ 2).

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.

5.11. Since Y_{1}, Y_{2} are uniformly distributed on this triangle of area 1, the probability that (Y_{1}, Y_{2}) lies in any region in the triangle is just the area of that region. (Careful – this is definitely false when the distribution is not uniform!)

(a) P (Y_{1} ≤ 3/4, Y_{2} ≤ 3/4 is the area of the remaining region when we “cut off” two small triangles of areas 1/32 and 1/16 from the big triangle. That leaves an area of 1 − 3/32 = 29/32.

(b) P (Y_{1} − Y_{2} ≥ 0). Y_{1} ≥ Y_{2} below the line y_{1} = y_{2} in the plane. This gives the triangle with vertices (0, 0), (1, 0), (1/2, 1/2), whose area is 1/4.

5.14. Given: f(y_{1}, y_{2}) = 6y_{1}^{2}y_{2} if 0 ≤ y_{1} ≤ y_{2}, 0 ≤ y_{1} + y_{2} ≤ 2 and 0 otherwise. The region where the density is nonzero is the interior of the triangle T with vertices (0, 0), (1, 1), (0, 2).

(a) We have f(y_{1}, y_{2}) ≥ 0 for all y_{1}, y_{2}, so the thing that needs to be checked is that the total integral is 1.

∫ ∫_{T} f(y_{1}, y_{2}) dA = ∫_{0}^{1} ∫_{y_{1}}^{2−y_{1}} 6y_{1}^{2}y_{2} dy_{2} dy_{1}

= ∫_{0}^{1} 3y_{1}^{2}((2 − y_{1})^{2} − y_{1}^{2}) dy_{1}

= ∫_{0}^{1} 12y_{1}^{2} − 12y_{1}^{3} dy_{1}

= 4y_{1}^{3} − 3y_{1}^{4}|_{0}^{1}

= 1

as required for a density function.

(b) Note that Y_{1} + Y_{2} = 1 defines a line that cuts through the triangle T from part (a). The region we want is the part of T that lies below that line:

P (Y_{1} + Y_{2} ≤ 1) = ∫_{0}^{1/2} ∫_{y_{1}}^{1−y_{1}} 6y_{1}^{2}y_{2} dy_{2} dy_{1}
= ∫_{0}^{1/2} 3y_{1}^{2}((1 − y_{1})^{2} − y_{1}^{2}) dy_{1}
= ∫_{0}^{1/2} 3y_{1}^{2} − 6y_{1}^{3} dy_{1}
= y_{1}^{3} − (3/2)y_{1}^{4}|_{0}^{1/2}
= 1/32.

5.15. The region where the joint density is nonzero is an infinite wedge-shaped region in the first quadrant bounded by the y_{1}-axis (y_{2} = 0) and the line y_{2} = y_{1}.

(a) The inequalities Y_{1} < 2 and Y_{2} > 1 define a small triangle within the wedge, with vertices at (1, 1), (2, 1), (2, 2). To get the desired probability here it is somewhat simpler to integrate first with respect to y_{1} (doing the other order leads to an integral by parts):

P (Y_{1} < 2, Y_{2} > 1) = ∫_{1}^{2} ∫_{y_{2}}^{2} e^{−y_{1}} dy_{1} dy_{2}
= ∫_{1}^{2} e^{−y_{2}} − e^{−2} dy_{2}
= e^{−1} − 2e^{−2}
≐ .0972.

(b) The inequality Y_{1} ≥ 2Y_{2} defines the lower half of the wedge-shaped region – that is, the region below the line Y_{2} = Y_{1}/2. Again, we set up the integrals to integrate with respedt to y_{1} first:

P (Y_{1} ≥ 2Y_{2}) = ∫_{0}^{∞} ∫_{2y_{2}}^{∞} e^{−y_{1}} dy_{1} dy_{2}

= ∫_{0}^{∞} e^{−2y_{2}} dy_{2}

= \frac{−e^{−2y_{2}}}{2} |_{0}^{∞}

= \frac{1}{2}.

(c) The inequality Y_{1} − Y_{2} ≥ 1 defines the portion of the wedge-shaped region below the line Y_{2} = Y_{1} − 1. This is a line parallel to the upper edge of the wedge, passing through (1, 0) on the Y_{1}-axis. The probability we want is

P (Y_{1} − Y_{2} ≥ 1) = ∫_{0}^{∞} ∫_{y_{2}+1}^{∞} e^{−y_{1}} dy_{1} dy_{2}
= ∫_{0}^{∞} e^{−y_{2}−1} dy_{2}
= −e^{−y_{2}−1}|_{0}^{∞}
= e^{−1}
≐ .3679.

Additional Problem

By the definition of the Γ function and the fact that the two integration variables are independent of each other (i.e. s is a constant with respect to t and vice versa),

Γ(p)Γ(q) = ∫_{0}^{∞} s^{p−1}e^{−s} ds · ∫_{0}^{∞} t^{q−1}e^{−t} dt
= ∫_{0}^{∞} s^{p−1}e^{−s} · ∫_{0}^{∞} t^{q−1}e^{−t} dt ds  (1)
= ∫_{0}^{∞} ∫_{0}^{∞} s^{p−1}t^{q−1}e^{−(s+t)} dt ds.

The region of integration here is the entire first quadrant in the (s, t)-plane (s= horizontal axis, t= vertical axis).

Now we do the first change of variables, letting u = s + t and putting u on the horizontal axis, s on the vertical axis in the new coordinate system. The s-axis in the old coordinate system maps to the line through (1, 1) in the new coordinates. The t-axis in the old coordinates maps to the u axis in the new coordinates and the first quadrant in the (s, t) plane maps to the “wedge” bounded by s = u and the u-axis in the new coordinates. This means that the new form of the double integral (1) is

∫_{0}^{∞} ∫_{0}^{u} s^{p−1}(u − s)^{q−1}e^{−u} ds du.  (2)

Next, we perform the second change of coordinates, letting λ = s/u, u = u. Note that the value of λ at any point is the slope of the line through that point and the origin. Hence the wedge corresponds to the infinite “rectangle” 0 ≤ λ ≤ 1, 0 ≤ u ≤ ∞ in the new coordinates, and with the given information about how the area element transforms, we get that (2) becomes:

∫_{0}^{∞} ∫_{0}^{1}(λu)^{p−1}(u − λu)^{q−1}e^{−u}u dλ du.  (3)

Then, working with the integrand in (3), we can simplify, then split the result back up into two one-variable integrals (reversing the process we did above in the first step):

∫_{0}^{∞} ∫_{0}^{1}(λu)^{p−1}(u − λu)^{q−1}e^{−u}u dλ du

= ∫_{0}^{∞} ∫_{0}^{1} λ^{p−1}(1 − λ)^{q−1}u^{p+q−1}e^{−u}dλ du
= ∫_{0}^{1} λ^{p−1}(1 − λ)^{q−1} dλ · ∫_{0}^{∞} u^{p+q−1}e^{−u} du
= B(p, q)Γ(p + q)

(by the definitions of the beta and gamma functions). We have shown that

Γ(p)Γ(q) = B(p, q)Γ(p + q),

from which we get the desired equality by dividing both sides by Γ(p + q). (Again, very cool!!)