Mathematics 375 – Probability and Statistics I Solutions – Problem Set 8 November 4, 2005

4.105/ a) 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 for 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)

b) 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}

4.107/ Using the formula of 4.106 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.

4.108/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.

4.110/ (Refer to the general formula for the mgf of a uniform random variable given inside the back cover of the text, and the result derived in Exercise 4.105.) a) m_{Y} (t) = \frac{e^{t}−1}{t}. b) With W = aY, m_{W} (t) = m_{Y} (at) = \frac{e^{at}−1}{at}. By uniqueness, this is the mgf for a uniformly distributed random variable on (0, a). c) Similarly, X = −aY ⇒ m_{X}(t) = m_{Y} (−at) = \frac{e^{−at}−1}{−at} = \frac{1−e^{−at}}{at}. This is the mgf for a uniformly distributed random variable on (−a, 0).

d) U = aY + b ⇒ m_{U}(t) = e^{bt}m_{Y} (at) = e^{bt} · \frac{e^{at}−1}{at} = \frac{e^{(a+b)t}−e^{bt}}{at}. This is the mgf for a uniformly distributed random variable on (b, a + b).

4.115/ 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.

4.120/ From 4.87, we have μ = E(L) = 276, σ = \sqrt{47664} ≐ 218.32. In Tchebysheff, the value of k for which 1 − 1/k^{2} = .89 is k = 3.015. Then (μ − 3.015σ, μ + 3.015σ) is the interval (−382.2, 934.24). (Since L is the loss due to downtime of the machine, negative values of loss mean positive profits.)

4.122/ a) Recall that the χ^{2} distribution with ν degrees of freedom is another name for the Gamma distribution with α = ν/2 and β = 2. (We will be using these a lot next semester, because they arise in understanding variances of samples from a normal distribution.) From our general formulas E(Y) = 7 and V (Y) = 14 (σ = \sqrt{14} ≐ 3.742). b) We could compute the corresponding probability numerically. For a “quick and dirty” estimate, though, we can also apply Tchebysheff: Note 23 = μ + 4.3σ. P (Y > μ + 4.3σ) ≤ P (|Y − μ| > 4.3σ) ≤ \frac{1}{(4.3)^{2}} ėq.054, which is quite small. And of course, this also includes P (Y < μ − 4.3σ).) So it is unlikely to find such a large value for Y.

5.3/ If Y_{1} is the number married, and Y_{2} is the number never married, then the joint probability function is

P (Y_{1} = y_{1}, Y_{2} = y_{2}) = \begin{matrix}\binom{4}{y_{1}} & \binom{3}{y_{2}} & \binom{2}{3−y_{1}−y_{2}} \\ ( & {9 \atop 3} & ) \end{matrix}

5.6/ a) We must have 1 = ∫_{0}^{1} ∫_{0}^{1} ky_{1}y_{2} dy_{2} dy_{1} = k/4. so k = 4. b) The joint cdf is

F (y_{1}, y_{2}) = P (Y_{1} ≤ y_{1}, Y_{2} ≤ y_{2}) = \{ \begin{matrix}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{matrix}

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.

5.9/ 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.

5.12/ Given f(y_{1}, y_{2}) = \{ \begin{matrix}6y_{1}^{2}y_{2} & if 0 ≤ y_{1} ≤ y_{2}, 0 ≤ y_{1} + y_{2} ≤ 2 \\ 0 & otherwise \end{matrix}. 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.

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),

(1)  \begin{aligned}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 \\ & = ∫_{0}^{∞} ∫_{0}^{∞} s^{p−1}t^{q−1}e^{−(s+t)} dt ds \end{aligned}

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:

( \begin{matrix}1 & 1 \\ 1 & 0 \end{matrix} ) · ( \begin{matrix}1 \\ 0 \end{matrix} ) = ( \begin{matrix}1 \\ 1 \end{matrix} ) ( \begin{matrix}1 & 1 \\ 1 & 0 \end{matrix} ) · ( \begin{matrix}0 \\ 1 \end{matrix} ) = ( \begin{matrix}1 \\ 0 \end{matrix} )

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

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

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:

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

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} λ^{p−1}(1 − λ)^{q−1}u^{p+q−1}e^{−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!!)