Mathematics 375 – Probability and Statistics 1 Solutions – Problem Set 8 November 14, 2003

4.104/a) By inspection, this is an exponential distribution with parameter β = θ, so by Example 4.13 in the text with α = 1, β = θ,

m(t) = \frac{1}{1 − θt}

(for t < \frac{1}{θ}). This could also be done, of course, by computing

m(t) = ∫_{0}^{∞} e^{ty} · e^{−y/θ} dy

directly.

b) We have

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

Similarly,

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

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

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.

4.109/This computation was also done in class. For a uniformly distributed Y on the interval (θ_{1}, θ_{2}),

\begin{aligned}m(t) & = ∫_{θ_{1}}^{θ_{2}} e^{ty} \frac{1}{θ_{2} − θ_{1}} dy \\ & = \frac{e^{θ_{2}t} − e^{θ_{1}t}}{(θ_{2} − θ_{1})t} \end{aligned}

4.113/Note that this is a “backwards” exponential density (nonzero for negative y). Part a) is a “warm-up” for computing the moment generating function (it’s the computation of the value m(3/2)):

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}

For part b), we do the general computation in exactly the same way, noting that t is constant with respect to the integration variable y:

m(t) = ∫_{−∞}^{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}

For c), m′(t) = \frac{−1}{(1+t)^{2}} , so m′(0) = E(Y) = 1. Then, m′′(t) = \frac{2}{(1+t)^{3}} , so m′′(0) = E(Y^{2}) = 2. Hence V (Y) = E(Y^{2}) − (E(Y))^{2} = 2 − 1 = 1.

4.114/Use Tchebysheff’s Theorem:

P (|Y − μ| ≤ kσ) ≥ 1 − \frac{1}{k^{2}}

To get an interval (μ − kσ, μ + kσ) that is guaranteed to include at least 90% of the tire lifetimes, we make 1 − \frac{1}{k^{2}} = .9, or k = \sqrt{10} ≐ 3.162. This gives

(μ − kσ, μ + kσ) = (25000 − (3.162)(4000), 25000 + (3.162)(4000)) = (12352, 37648)

4.116/To apply Tchebysheff’s Theorem, we need to know μ and σ, so we begin by computing those values. With the given density function,

μ = E(Y) = ∫_{0}^{2} y · \frac{1}{2}(2 − y) dy

= (\frac{y^{2}}{2} − \frac{y^{3}}{6} |_{0}^{2}

= \frac{2}{3}

Then

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

= (\frac{y^{3}}{3} − \frac{y^{4}}{8} |_{0}^{2}

= \frac{2}{3}

So σ = \sqrt{(2/3) − (2/3)^{2}} = \sqrt{2}/3. In this case, μ − 2σ = (2 − 2\sqrt{2})/3 < 0, so

P (|Y − μ| < 2σ) = ∫_{0}^{(2+2\sqrt{2})/3} \frac{1}{2}(2 − y) dy ≐ .96

Compared with the empirical rule value .95 and the Tchebysheff lower bound 1 − \frac{1}{2^{2}} = .75, we see that the empirical rule is much closer.

4.117/From previous work we know that the mean and standard deviation for the uniform distribution on (a, b) are μ = \frac{a+b}{2} (the midpoint of the interval), and σ = \frac{b−a}{2\sqrt{3}}, which equals \frac{1}{\sqrt{3}} times the distance from the midpoint to either endpoint. In this case

μ + 2σ = (\frac{a + b}{2}) + 2 · (\frac{b − a}{2\sqrt{3}}) = (\frac{a + b}{2}) + \frac{2}{\sqrt{3}} · (\frac{b − a}{2})

Since 2 > \sqrt{3}, this takes us past the right endpoint of the interval. Similarly, μ − 2σ takes us past the left endpoint. But the uniform density is only nonzero between a and b, so

P (|Y − μ| ≤ 2σ) = ∫_{a}^{b} \frac{1}{b − a} dy = 1

Compared with the empirical rule value .95 and the Tchebysheff lower bound 1 − \frac{1}{2^{2}} = .75, we again see that the empirical rule is much closer, although it is actually an underestimate in this case(!).

4.118/From previous work, we know that the exponential distribution with parameter β has mean μ = β, and standard deviation σ = β. The interval (μ − 2σ, μ + 2σ) = (−β, 3β) extends past the left endpoint of the interval where the density is nonzero. Hence

P (|Y − μ| ≤ 2σ) = ∫_{0}^{3β} e^{−y/β} \frac{1}{β} dy = 1 − e^{−3} ≐ .9502

Compared with the empirical rule value .95 and the Tchebysheff lower bound 1 − \frac{1}{2^{2}} = .75, we again see that the empirical rule is much closer.

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!!)