Measure arrival times in hours after noon, and let Alice’s and Bob’s arrival times be \[ X,Y\sim \operatorname {Unif}(0,1), \] independently. They meet exactly when their arrival times differ by at most 15 minutes, which is one quarter of an hour. Thus the desired event is \[ |X-Y|\le \frac 14. \]

Geometrically, the pair \((X,Y)\) is uniformly distributed over the unit square. The event \(|X-Y|>1/4\) consists of two congruent right triangles, one above the line \(y=x+1/4\) and one below the line \(y=x-1/4\). Each triangle has side length \(3/4\), so the total excluded area is \[ 2\cdot \frac 12\left (\frac 34\right )^2=\frac {9}{16}. \] Therefore the desired probability is \[ 1-\frac {9}{16}=\frac {7}{16}. \]

(a) Since \(X\sim \operatorname {Unif}(0,L)\), we have \[ f_X(x)=\frac 1L,\qquad 0<x<L. \] Given \(X=x\), the random variable \(Y\) is uniform on \((0,x)\), so \[ f_{Y\mid X}(y\mid x)=\frac 1x,\qquad 0<y<x. \] Thus \[ f_{X,Y}(x,y)=f_X(x)f_{Y\mid X}(y\mid x)=\frac {1}{Lx}, \qquad 0<y<x<L. \] It is zero outside this region.

(b) Marginalizing out \(Y\), for \(0<x<L\), gives \[ f_X(x)=\int _0^x \frac {1}{Lx}\,dy=\frac {1}{Lx}\cdot x=\frac 1L. \] This is exactly the \(\operatorname {Unif}(0,L)\) density.

(c) Using the definition of a conditional density, \[ f_{Y\mid X}(y\mid x)=\frac {f_{X,Y}(x,y)}{f_X(x)} =\frac {1/(Lx)}{1/L}=\frac 1x, \qquad 0<y<x. \] This is the uniform density on \((0,x)\), as expected.

(d) For a fixed \(y\), the possible values of \(x\) run from \(y\) to \(L\). Hence, for \(0<y<L\), \[ f_Y(y)=\int _y^L \frac {1}{Lx}\,dx =\frac 1L\log \left (\frac {L}{y}\right ). \]

(e) Finally, \[ f_{X\mid Y}(x\mid y) =\frac {f_{X,Y}(x,y)}{f_Y(y)} =\frac {1/(Lx)}{(1/L)\log (L/y)} =\frac {1}{x\log (L/y)}, \] for \(y<x<L\).

(a) Scale the stick to have length 1. Let the two breakpoints be independent uniform points on \((0,1)\), and order them as \(U<V\). The three piece lengths are \[ U, \qquad V-U, \qquad 1-V. \] A triangle can be formed exactly when no piece has length at least \(1/2\). The ordered pair \((U,V)\) is uniform over the triangle \[ 0<U<V<1. \] Inside this triangle, the forbidden regions are \[ U>\frac 12, \qquad V-U>\frac 12, \qquad 1-V>\frac 12. \] These three regions are congruent corner triangles whose total area is three quarters of the area of the full region. Therefore the remaining area is one quarter of the full region.

(b) The table stands exactly when the three points on the circumference are not all contained in some semicircle. Equivalently, the triangle formed by the three leg positions contains the center of the table.

For three independent uniform points on a circle, the probability that they all lie in some semicircle is \(3/4\). Thus the complementary probability, namely the probability that the table stands, is \[ 1-\frac 34=\frac 14. \]

The joint CDF is \[ F(x,y)=P(X\le x,\,Y\le y). \] We want the probability of the rectangle \[ a_1\le X\le a_2, \qquad b_1\le Y\le b_2. \] Start with the probability of the lower-left rectangle ending at \((a_2,b_2)\), namely \(F(a_2,b_2)\). This includes too much. It includes points with \(X\le a_1\), whose probability is \(F(a_1,b_2)\), and points with \(Y\le b_1\), whose probability is \(F(a_2,b_1)\). After subtracting these two pieces, the corner \((X\le a_1,Y\le b_1)\) has been subtracted twice, so it must be added back once. Thus \[ P(a_1\le X\le a_2,\,b_1\le Y\le b_2) = F(a_2,b_2)-F(a_1,b_2)-F(a_2,b_1)+F(a_1,b_1). \] This is the same expression as the one stated in the problem, only with the last two terms written in the usual order.

(a) The function is nonnegative on the unit square. Its total integral is \[ \int _0^1\int _0^1 (x+y)\,dy\,dx = \int _0^1\left (x+\frac 12\right )dx =\frac 12+\frac 12=1. \] Thus it is a valid joint PDF.

(b) The variables are not independent. One way to see this is to compute the marginals and observe that their product does not equal the joint density.

(c) For \(0<x<1\), \[ f_X(x)=\int _0^1 (x+y)\,dy=x+\frac 12. \] Similarly, for \(0<y<1\), \[ f_Y(y)=\int _0^1 (x+y)\,dx=y+\frac 12. \]

(d) For \(0<x<1\), the conditional density is \[ f_{Y\mid X}(y\mid x) =\frac {f_{X,Y}(x,y)}{f_X(x)} =\frac {x+y}{x+1/2}, \qquad 0<y<1. \]

(a) We need \[ \int _0^1\int _0^y cxy\,dx\,dy=1. \] The integral is \[ c\int _0^1 y\left (\int _0^y x\,dx\right )dy = c\int _0^1 y\cdot \frac {y^2}{2}\,dy = \frac {c}{2}\cdot \frac 14 = \frac {c}{8}. \] Thus \(c=8\).

(b) The variables are not independent. The support itself already shows this, since the condition \(0<X<Y<1\) ties the possible values of \(X\) and \(Y\) together.

(c) With \(c=8\), for \(0<x<1\), \[ f_X(x)=\int _x^1 8xy\,dy=4x(1-x^2). \] For \(0<y<1\), \[ f_Y(y)=\int _0^y 8xy\,dx=4y^3. \]

(d) For fixed \(x\), the conditional support is \(x<y<1\). Therefore \[ f_{Y\mid X}(y\mid x) =\frac {8xy}{4x(1-x^2)} =\frac {2y}{1-x^2}, \qquad x<y<1. \]

The polynomial has a real root exactly when its discriminant is nonnegative. Here that condition is \[ B^2-4A\ge 0, \] or equivalently \[ A\le \frac {B^2}{4}. \] Since \((A,B)\) is uniformly distributed on the unit square, the desired probability is the area of the region under the curve \(a=b^2/4\), for \(0<b<1\). Hence \[ P\left (A\le \frac {B^2}{4}\right ) = \int _0^1 \frac {b^2}{4}\,db = \frac {1}{4}\cdot \frac 13 = \frac {1}{12}. \]

Let \[ D=|X-Y|. \] We need \(\operatorname {SD}(D)\). First, \[ E(D^2)=E\big ((X-Y)^2\big )=\operatorname {Var}(X-Y). \] Since \(X\) and \(Y\) are independent uniforms, \[ \operatorname {Var}(X-Y)=\operatorname {Var}(X)+\operatorname {Var}(Y)=\frac {1}{12}+\frac {1}{12}=\frac 16. \] Also, by the usual triangular-density calculation for the distance between two independent uniforms, \[ E(D)=\int _0^1 2d(1-d)\,dd=\frac 13. \] Therefore \[ \operatorname {Var}(D) =E(D^2)-[E(D)]^2 =\frac 16-\frac 19 =\frac {1}{18}. \] Thus \[ \operatorname {SD}(D)=\sqrt {\frac 1{18}}=\frac {1}{3\sqrt 2}. \]

Let \(q=1-p\), and use the convention that \(\operatorname {Geom}(p)\) counts the number of failures before the first success, so \[ P(X=k)=q^k p, \qquad k=0,1,2,\ldots . \]

(a) If \(0\le l<m\), then one of \(X,Y\) must equal \(l\) and the other must equal \(m\), so \[ P(L=l,M=m)=2p^2q^{l+m}. \] If \(l=m\), then both variables must equal \(l\), so \[ P(L=l,M=l)=p^2q^{2l}. \] The variables \(L\) and \(M\) are not independent. For example, the event \(M<L\) has probability zero, so knowing \(L\) restricts possible values of \(M\).

(b) Using the joint PMF, \[ P(L=l)=p^2q^{2l}+\sum _{m=l+1}^{\infty }2p^2q^{l+m}. \] The geometric sum gives \[ P(L=l)=p^2q^{2l}+2p^2q^l\frac {q^{l+1}}{1-q} =p^2q^{2l}+2pq^{2l+1} =p(2-p)q^{2l}. \]

The story proof is even cleaner. The event \(L=l\) means that during the first \(l\) paired trials, both sequences fail, and at time \(l\), at least one of them succeeds. The probability of both failing in a paired trial is \(q^2\), and the probability that at least one succeeds is \(1-q^2=p(2-p)\). Hence \[ P(L=l)=(q^2)^l(1-q^2)=p(2-p)q^{2l}. \]

(c) Since \[ L+M=X+Y, \] we have \[ E(M)=E(X)+E(Y)-E(L). \] For our convention, \(E(X)=E(Y)=q/p\). Also, from part (b), \[ E(L)=\frac {q^2}{1-q^2}=\frac {q^2}{p(2-p)}. \] Therefore \[ E(M)=\frac {2q}{p}-\frac {q^2}{p(2-p)}. \]

(d) Let \(D=M-L\). If \(d=0\), then \(M=L=l\), so \[ P(L=l,D=0)=p^2q^{2l}. \] If \(d>0\), then \((X,Y)\) can be \((l,l+d)\) or \((l+d,l)\), so \[ P(L=l,D=d)=2p^2q^{2l+d}. \] The factorization shows that \(L\) and \(D=M-L\) are independent. Indeed, \[ P(L=l)=p(2-p)q^{2l}, \] while \[ P(D=0)=\frac {p}{2-p}, \qquad P(D=d)=\frac {2p}{2-p}q^d,\quad d=1,2,\ldots , \] and multiplying these marginals recovers the joint PMF above.

(a) Since the two dice are independent and identically distributed, \[ \operatorname {Cov}(X+Y,X-Y) = \operatorname {Cov}(X,X)-\operatorname {Cov}(X,Y) +\operatorname {Cov}(Y,X)-\operatorname {Cov}(Y,Y). \] The cross-covariances are zero, and \(\operatorname {Var}(X)=\operatorname {Var}(Y)\). Therefore \[ \operatorname {Cov}(X+Y,X-Y)=\operatorname {Var}(X)-\operatorname {Var}(Y)=0. \] They are not independent. For instance, if \(X+Y=2\), then necessarily \(X=Y=1\), so \(X-Y=0\). Knowing the sum can restrict the difference.

(b) The same covariance calculation applies for independent uniforms. We again get \[ \operatorname {Cov}(X+Y,X-Y)=\operatorname {Var}(X)-\operatorname {Var}(Y)=0. \] They are not independent. The pair \((X+Y,X-Y)\) cannot take arbitrary values in a rectangle. For example, if \(X+Y\) is very close to 0, then both \(X\) and \(Y\) must be close to 0, forcing \(X-Y\) also to be close to 0.

Think of the multinomial vector as coming from \(n\) independent trials, where each trial falls into exactly one of the \(k\) categories. For trial \(r\), define \[ I_{r,i}=\mathbf {1}\{\text {trial }r\text { falls in category }i\}. \] Then \[ X_i=\sum _{r=1}^n I_{r,i}, \qquad X_j=\sum _{s=1}^n I_{s,j}. \] Therefore \[ \operatorname {Cov}(X_i,X_j) = \sum _{r=1}^n\sum _{s=1}^n \operatorname {Cov}(I_{r,i},I_{s,j}). \] When \(r\ne s\), the indicators come from different independent trials, so their covariance is zero. When \(r=s\), the two indicators cannot both be 1 because a single trial cannot fall into both category \(i\) and category \(j\). Hence \[ E(I_{r,i}I_{r,j})=0, \] while \[ E(I_{r,i})=p_i, \qquad E(I_{r,j})=p_j. \] Thus \[ \operatorname {Cov}(I_{r,i},I_{r,j}) =0-p_ip_j =-p_ip_j. \] There are \(n\) such same-trial terms, so \[ \operatorname {Cov}(X_i,X_j)=n(-p_ip_j)=-np_ip_j. \]