Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 9 November 12, 2009

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.

5.25. (a) The marginal densities are

f_{1}(y_{1}) = ∫_{0}^{∞} e^{−y_{1}−y_{2}} dy_{2} = −e^{−y_{1}−y_{2}} |_{0}^{∞} = e^{−y_{1}}

if y_{1} > 0 and zero otherwise. Similarly,

f_{2}(y_{2}) = ∫_{0}^{∞} e^{−y_{1}−y_{2}} dy_{1} = −e^{−y_{1}−y_{2}} |_{0}^{∞} = e^{−y_{2}}

if y_{2} > 0 and zero otherwise. These are both exponential densities with β = 1.

(b) P (1 < Y_{i} < 2.5) = ∫_{1}^{2.5} e^{−y_{i}} dy_{i} = e^{−1} − e^{−2.5} for i = 1, 2.

(c) The conditional density is only defined for y_{2} > 0 (since we must have f_{2}(y_{2}) > 0 in Definition 5.7).

(d) The conditional density f(y_{1}|y_{2}) = \frac{e^{−y_{1}−y_{2}}}{e^{−y_{2}}} = e^{−y_{1}}.

(e) Similarly, f(y_{1}|y_{2}) = e^{−y_{2}}.

(f) The conditional and marginal densities for both variables are equal.

(g) The marginal and conditional probabilities in any interval will also be equal.

5.26. (a) The marginal densities are

f_{1}(y_{1}) = ∫_{0}^{1} 4y_{1}y_{2} dy_{2} = 2y_{1}, and similarly
f_{2}(y_{2}) = ∫_{0}^{1} 4y_{1}y_{2} dy_{1} = 2y_{2}.

Looking ahead to the next section, note that these satisfy the criterion of Theorem 5.4. So Y_{1} and Y_{2} are independent.

(b) To compute the denominator in the conditional probability, we can use the marginal density from part a:

P (Y_{1} ≤ 1/2|Y_{2} ≥ 3/4) = \frac{P (Y_{1} ≤ 1/2, Y_{2} ≥ 3/4)}{P (Y_{2} ≥ 3/4)}

= \frac{∫_{0}^{1/2} ∫_{3/4}^{1} 4y_{1}y_{2} dy_{2} dy_{1}}{∫_{3/4}^{1} 2y_{2} dy_{2}}

= \frac{∫_{0}^{1/2} 2y_{1} dy_{1} ∫_{3/4}^{1} 2y_{2} dy_{2}}{∫_{3/4}^{1} 2y_{2} dy_{2}}

= ∫_{0}^{1/2} 2y_{1} dy_{1}

= 1/4.

(The cancellation works this way because of independence.)

5.54. No – Y_{1}, Y_{2} cannot be independent when the interval of y_{2} values where the density is nonzero depends on y_{1}. (You can also see this directly by computing f_{1}(y_{1}) and f_{2}(y_{2}). The region where the joint density is nonzero is the triangle with vertices at (0, 0), (2, 0), (2, 1). So

f_{1}(y_{1}) = ∫_{0}^{y_{1}/2} 1 dy_{2} = \frac{y_{1}}{2}

(when 0 ≤ y_{1} ≤ 2; 0 otherwise). Similarly,

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

(when 0 ≤ y_{2} ≤ 1; 0 otherwise). The product f_{1}(y_{1})f_{2}(y_{2}) is not the same as f(y_{1}, y_{2}).)

5.59. No – same reason as 5.54. The marginal densities are

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

and

f_{2}(y_{2}) = ∫_{y_{2}}^{∞} e^{−y_{1}} dy_{1} = e^{−y_{2}}.

Note that f(y_{1}, y_{2}) ≠ f_{1}(y_{1})f_{2}(y_{2}).

5.65. (a) The marginal density of Y_{1} is computed as follows. Note that we try to split up the integral so that the integrand is a product of functions of y_{1} and y_{2} alone. That allows factors depending only on y_{1} to be taken outside the integral sign:

f_{1}(y_{1}) = ∫_{0}^{∞} (1 − α(1 − 2e^{−y_{1}})(1 − 2e^{−y_{2}})) e^{−y_{1}−y_{2}} dy_{2}
= e^{−y_{1}} ∫_{0}^{∞} e^{−y_{2}} dy_{2} − α(1 − 2e^{−y_{1}})e^{−y_{1}} ∫_{0}^{∞}(1 − 2e^{−y_{2}})e^{−y_{2}} dy_{2}
= e^{−y_{1}} ∫_{0}^{∞} e^{−y_{2}} dy_{2} − −α(1 − 2e^{−y_{1}})e^{−y_{1}} · (∫_{0}^{∞} e^{−y_{2}} dy_{2} − ∫_{0}^{∞} 2e^{−2y_{2}} dy_{2})
= e^{−y_{1}} · 1 − α(1 − 2e^{−y_{1}})e^{−y_{1}}(1 − 1)
= e^{−y_{1}}

(since the integrals in the next to last line are all integrals of exponential densities over the interval where they are nonzero, hence each equals 1). Hence Y_{1} is exponentially distributed with mean β = 1.

(b) We have f_{2}(y_{2}) = e^{−y_{2}} by the symmetry of the joint density. So Y_{2} is also exponential with mean β = 1.

(c) By a general result (Theorem 5.4), Y_{1} and Y_{2} are independent if and only if f(y_{1}, y_{2}) = f_{1}(y_{1})f_{2}(y_{2}). Here, this is true if and only if

e^{−y_{1}−y_{2}} = (1 − α(1 − 2e^{−y_{1}})(1 − 2e^{−y_{2}})) e^{−y_{1}−y_{2}} = e^{−y_{1}−y_{2}} − α(1 − 2e^{−y_{1}})(1 − 2e^{−y_{2}})e^{−y_{1}−y_{2}}

which is true if and only if α = 0.

5.76. This can be done most easily by using the marginal densities for the Y_{i} (although there are several other valid methods too). Note that f_{1}(y_{1}) = ∫_{0}^{1} 4y_{1}y_{2} dy_{2} = 2y_{1} if 0 ≤ y_{1} ≤ 1, and zero otherwise. Similarly for f_{2}(y_{2}) by the symmetry of the joint density.

(a) E(Y_{1}) = ∫_{0}^{1} y_{1}f_{1}(y_{1}) dy_{1} = ∫_{0}^{1} 2y_{1}^{2} dy_{1} = \frac{2}{3}.

(b) Then the variance is:

V (Y_{1}) = E(Y_{1}^{2}) − (\frac{2}{3})^{2} = ∫_{0}^{1} 2y_{1}^{3} dy_{1} − \frac{4}{9} = \frac{1}{2} − \frac{4}{9} = \frac{1}{18}.

(c) By linearity, E(Y_{1} − 3Y_{2}) = E(Y_{1}) − 3E(Y_{2}) = \frac{2}{3} − 2 = \frac{−4}{3}. This can also be computed, for instance, as the integral:

∫_{0}^{1} ∫_{0}^{1}(y_{1} − 3y_{2}) · 4y_{1}y_{2} dy_{2} dy_{1}.

5.79. To integrate over this region, is probably easiest to integrate with respect to y_{1} first:

E(Y_{1}Y_{2}) = ∫_{0}^{1} ∫_{y_{2}−1}^{−y_{2}+1} y_{1}y_{2} dy_{1} dy_{2}

= ∫_{0}^{1} y_{2} \frac{(−y_{2} + 1)^{2} − (y_{2} − 1)^{2}}{2} dy_{1}

= 0,

since everything cancels out in the integrand.

5.80. Using the joint density, we multiply out and integrate term by term:

E(30Y_{1} + 25Y_{2}) = ∫_{0}^{1} ∫_{0}^{1}(30y_{1} + 25y_{2})(y_{1} + y_{2}) dy_{2} dy_{1}

= ∫_{0}^{1} ∫_{0}^{1} 30y_{1}^{2} + 55y_{1}y_{2} + 25y_{2}^{2} dy_{2} dy_{1}

= ∫_{0}^{1} 30y_{1}^{2} + \frac{55}{2}y_{1} + \frac{25}{3} dy_{1}

= \frac{385}{12}.

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}udλ 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} 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!!)