PROB_AS1

Vasudeva S N

1

1 a

We know that $A\subseteq B⟹P(A)\le P(B)$. Thus, $P(A\cap B\cap C)\le P(A)$, $P(A\cap B\cap C)\le P(B)$, $P(A\cap B\cap C)\le P(C)$. It follows that $P(A\cap B\cap C)\le P(A)\wedge P(B)\wedge P(C)$.

1 b

Similarly to part a, we have $P(A\cup B\cup C)\ge P(A)$, $P(A\cup B\cup C)\ge P(B)$, $P(A\cup B\cup C)\ge P(C)$. It follows that $P(A\cup B\cup C)\ge P(A)\vee P(B)\vee P(C)$.

1 c

Let $C=(A\cup B{)}^c$. Then,

$$\begin{array}{rl}P(A\cup B\cup C)=1& =P(A)+P(B)+P(C)-P(A\cap B)\\ & \ge P(A)+P(B)-P(A\cap B)\end{array}$$

Therefore, $P(A\cap B)\ge P(A)+P(B)-1$.

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2

2 a

For each draw of $r$ balls, let $X$ denote the maximum number drawn. Then, $P(X\le m)={\left(\frac{m}{n}\right)}^r$. Therefore, $P(X=m)=P(X\le m)-P(X\le m-1)={\left(\frac{m}{n}\right)}^r-{\left(\frac{m-1}{n}\right)}^r$.

2 b

For each draw of $r$ balls, let $X$ denote the maximum number drawn. Then, $P(X\le m)=\left(\genfrac{}{}{0ex}{}{m}{r}\right){\left(\genfrac{}{}{0ex}{}{n}{r}\right)}^{-1}$. Therefore, $P(X=m)=P(X\le m)-P(X\le m-1)=$

$$\begin{array}{r}\frac{\left(\genfrac{}{}{0ex}{}{m}{r}\right)-\left(\genfrac{}{}{0ex}{}{m-1}{r}\right)}{\left(\genfrac{}{}{0ex}{}{n}{r}\right)}=\frac{\left(\genfrac{}{}{0ex}{}{m-1}{r-1}\right)}{\left(\genfrac{}{}{0ex}{}{n}{r}\right)}.\end{array}$$

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3

$$\begin{array}{rl}P(\text{A wins})& =\sum _{i=1}^{\infty }P(\text{A wins in round }i)\\ & =\sum _{i=1}^{\infty }\frac{5}{36}{\left(\frac{31}{36}\frac{5}{6}\right)}^{i-1}\\ & =\frac{30}{61}\end{array}$$

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4

Let $WB$ represent the event that a write ball is picked in the first draw, and a black ball is picked in the second draw. $WW,BB,BW$ are also events, and these four and disjoint and exhaustive. Let $A$ be the event that a white ball was picked in the third draw. Then,

$$\begin{array}{rl}P(A)=& P(A|WW)P(WW)+P(A|WB)P(WB)+\\ & P(A|BW)P(BW)+P(A|BB)P(BB)\\ =& \frac{1}{w_1+b_1}\frac{1}{w_1+w_2+1}\frac{1}{w_1+b_1}\\ & \left(w_1(w_1+1)w_1+(w_1-1)w_2{w}_1+(w_1+1)w_1{b}_1+w_1(w_2+1)b_1\right)\\ & \\ =& \frac{w_1\left((w_1+1)w_1+(w_1-1)w_2+(w_1+1)b_1+(w_2+1)b_1\right)}{(w_1+b_1{)}^2(w_1+w_2+1)}.\end{array}$$

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5

The corrected version of the question defines $\sim$ by $X\sim Y$ if $P(X=n)=P(Y=n)\forall n\in \mathbb{Z}$, $i$.$e$, $X$ and $Y$ have the same PMF. $d([X],[Y])$ is defined by

$$d([X],[Y])=\sum _{n\in \mathbb{Z}}\left|P(X=n)-P(Y=n)\right|.$$

Clearly, $d([X],[Y])\ge 0$. If $d([X],[Y])=0$, $P(X=n)=P(Y=n)\forall n\in \mathbb{Z}$, so $[X]=[Y]$. On the other hand, if $[X]=[Y]$, $d([X],[Y])=0$. Thus, $d$ is positive definite. Clearly, $d$ is symmetric and reflexive. If $[X],[Y],[Z]\in \widetilde{Ł}$, then

$$\begin{array}{rl}d([X],[Y])& =\sum _{n\in \mathbb{Z}}|P(X=n)-P(Y=n)|\\ & \le \sum _{n\in \mathbb{Z}}|P(X=n)-P(Z=n)|+\sum _{n\in \mathbb{Z}}|P(Z=n)-P(Y=n)|\\ & =d([X],[Z])+d([Z],[Y]).\end{array}$$

The triangle inequality holds. Therefore, $d$ is a metric on $\widetilde{Ł}$.

Fix $[X]$, $[Y]$, and their respective representatives $X$ and $Y$. Let $\mathbf{\alpha }\subset \mathbb{Z}$ be the set of all integers $a$ for which $P(X=a)\ge P(Y=a)$. Then,

$$\begin{array}{rl}0& =P(X\in \mathbb{Z})-P(Y\in \mathbb{Z})\\ & =\sum _{n\in \mathbb{Z}}P(X=n)-P(Y=n)\\ & =\sum _{n\in \mathbf{\alpha }}P(X=n)-P(Y=n)+\sum _{n\in {\mathbf{\alpha }}^c}P(X=n)-P(Y=n).\end{array}$$

Next,

$$\begin{array}{rl}d([X],[Y])& =\sum _{n\in \mathbb{Z}}\left|P(X=n)-P(Y=n)\right|\\ & =\sum _{n\in \mathbf{\alpha }}P(X=n)-P(Y=n)-\sum _{n\in {\mathbf{\alpha }}^c}P(X=n)-P(Y=n)\\ & =2\sum _{n\in \mathbf{\alpha }}P(X=n)-P(Y=n).\end{array}$$

Now, for some $A\subset \mathbb{Z}$,

$$\begin{array}{rl}& |P(X\in A)-P(Y\in A)|\\ & =\left|\sum _{n\in A\cap \mathbf{\alpha }}P(X=n)-P(Y=n)+\sum _{n\in A\cap {\mathbf{\alpha }}^c}P(X=n)-P(Y=n)\right|\\ & \le \frac{d([X],[Y])}{2}.\end{array}$$

Thus, $d([X],[Y])/2$ is an upper bound for $|P(X\in A)-P(Y\in A)|$. Next, we will show that no real number less than $d([X],[Y])/2$ is an upper bound. Let $ϵ>0$. Let ${\mathbf{\alpha }}_1^c,{\mathbf{\alpha }}_2^c,\dots$ be an enumeration of the elements in ${\mathbf{\alpha }}^c$.

$$\sum _{n=1}^{\infty }P(X={\mathbf{\alpha }}_n^c)-P(Y={\mathbf{\alpha }}_n^c)=-\frac{d([X],[Y])}{2}.$$

Each term in the above series is negative. Further, the limit as $n\to \infty$ of individual terms of a convergent series must be zero. Thus, there exists $n$ such that $|P(X={\mathbf{\alpha }}_n^c)-P(Y={\alpha }_n^c)|<ϵ$. Now, let $A$ be $\mathbf{\alpha }\cup \{n\}$. Then,

$$\begin{array}{r}|P(X\in A)-P(Y\in A)|>\left|\frac{d([X],[Y])}{2}-ϵ\right|\end{array}$$

Thus, $d([X],[Y])/2$ is the supremum of $|P(X\in A)-P(Y\in A)|$, $A\subseteq \mathbb{Z}$.

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6

Let $X\sim \text{Pois}(\lambda )$. Then, $f_X(x)$ is given by $\frac{{\lambda }^x}{x!}{e}^{-\lambda }$ for $x=0,1,2,\dots$. Consider the ratio $\frac{f_X(x)}{f_X(x-1)}=\frac{\lambda }{x}$. This ratio decreases with $x$, and $f_X(x)>f_X(x-1)$ only if the ratio is greater than 1. We can therefore conclude that

$$\cdots <f_X(\lfloor \lambda \rfloor -1)<f_X(\lfloor \lambda \rfloor )>f_X(\lfloor \lambda \rfloor +1)>\dots ,$$

that is, the most probable value is $\lfloor \lambda \rfloor$.

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7

Part a

The number of ways to obtain a sum of $k$ with $n$ throws is given by

$$\sum _{i=0}^n(-1{)}^i(\genfrac{}{}{0ex}{}{n}{i}\left)\right(\genfrac{}{}{0ex}{}{k-6i-1}{n-1})$$

The total number of ways of obtaining sum $k$ is given by

$$\sum _{n=1}^k\sum _{i=0}^n(-1{)}^i(\genfrac{}{}{0ex}{}{n}{i}\left)\right(\genfrac{}{}{0ex}{}{k-6i-1}{n-1})$$

The total probability of obtaining sum $k$ is given by

$$\sum _{n=1}^k{\left(\frac{1}{12}\right)}^n\sum _{i=0}^n(-1{)}^i(\genfrac{}{}{0ex}{}{n}{i}\left)\right(\genfrac{}{}{0ex}{}{k-6i-1}{n-1})$$

Thus,

$$\begin{array}{rl}P(N=2|S=4)& =\frac{{\left(\frac{1}{12}\right)}^2\sum _{i=0}^2(-1{)}^i(\genfrac{}{}{0ex}{}{2}{i}\left)\right(\genfrac{}{}{0ex}{}{4-6i-1}{2-1})}{\sum _{n=1}^4{\left(\frac{1}{12}\right)}^n\sum _{i=0}^n(-1{)}^i(\genfrac{}{}{0ex}{}{n}{i}\left)\right(\genfrac{}{}{0ex}{}{4-6i-1}{n-1})}\\ & =0.196631770596268\end{array}$$

Part b

Clearly, $P(N\text{ is even})=\frac{1}{2}$. Also,

$$\begin{array}{rl}P(\{S=4\}\cap \{N\text{ is even }\})& =\sum _{n\in \{2,4\}}{\left(\frac{1}{12}\right)}^n\sum _{i=0}^n(-1{)}^i(\genfrac{}{}{0ex}{}{n}{i}\left)\right(\genfrac{}{}{0ex}{}{4-6i-1}{n-1})\\ & =0.0208815586419753\end{array}$$

Thus, $P(S=4|N\text{ is even})=0.0417631172839506$.

Part c

At this point I’ll just ditch the formula. If $A$ is the given event,

$$\begin{array}{rl}P(A)& =\frac{P(1,3)}{P(1,3)+P(1,1,1)+P(1,2,1)+P(1,1,1,1)}\\ & =\frac{{\left(\frac{1}{12}\right)}^2}{{\left(\frac{1}{12}\right)}^2+2{\left(\frac{1}{12}\right)}^3+{\left(\frac{1}{12}\right)}^4}\\ & =0.852071005917\end{array}$$

Part d

Let $X$ denote the largest number shown by a die. Then,

$$p(X\le r)=\sum _{n=1}^{\infty }{\left(\frac{r}{12}\right)}^n=\frac{r}{12-r}$$

It follows that

$$P(X=r)=P(X\le r)-P(X\le r-1)=\frac{12}{(12-r)(13-r)}.$$

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8

Part i

$f_X(k)=f_Y(k)=p(1-p{)}^{k-1}$.

$$\begin{array}{rl}P(Y>X)& =\sum _{x=1}^{\infty }P(Y>x|X=x)P(X=x)\\ & =\sum _{x=1}^{\infty }(1-p{)}^{x}p(1-p{)}^{x-1}\\ & =\frac{p}{1-p}\sum _{x=1}^{\infty }(1-p{)}^{2x}\\ & =\frac{1-p}{2-p}\end{array}$$

Part ii

$$\begin{array}{rl}P(X=Y)& =\sum _{x=1}^{\infty }P(Y=x|X=x)P(X=x)\\ & =p^2\sum _{x=1}^{\infty }(1-p{)}^{2x-2}\\ & =\frac{p}{2-p}\end{array}$$

Part iii

$$\begin{array}{rl}P(X\ge kY)& =\sum _{y=1}^{\infty }P(X\ge ky|Y=y)P(Y=y)\\ & =\sum _{y=1}^{\infty }(1-p{)}^{ky-1}p(1-p{)}^{y-1}\\ & =p\sum _{y=1}^{\infty }(1-p{)}^{ky+y-2}\\ & =\frac{p(1-p{)}^{k-1}}{1-(1-p{)}^{k+1}}\end{array}$$

Part iv

$$\begin{array}{rl}P(X\text{ divides }Y)& =\sum _{x=1}^{\infty }P(x\text{ divides }Y|X=x)P(X=x)\\ & =\sum _{x=1}^{\infty }P(Y=kx,k\in \mathbb{N})p(1-p{)}^{x-1}\\ & =p^2\sum _{x=1}^{\infty }\left(\sum _{k\ge 1}(1-p{)}^{kx-1}\right)(1-p{)}^{x-1}\\ & =p^2\sum _{x=1}^{\infty }\frac{(1-p{)}^{2x-2}}{1-(1-p{)}^x}\end{array}$$

Part v

Let $m$ be the smallest integer such that $rm$ is an integer. Then,

$$\begin{array}{rl}P(X=rY)& =\sum _{y=1}^{\infty }P(X=ry|Y=y)P(Y=y)\\ & =\sum _{y=1}^{\infty }P(X=mry|Y=my)P(Y=my)\\ & =\frac{p^2}{(1-p{)}^2}\sum _{y=1}^{\infty }(1-p{)}^{my(r+1)}\\ & =\frac{p^2(1-p{)}^{mr+m-2}}{1-(1-p{)}^{mr+m}}.\end{array}$$

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9

The probability that the elevator does not stop any any level less than or equal to $k$ is given by ${\left(\frac{f-k}{k}\right)}^N$. Thus, the probability that the elevator will stop for the first time at level $k$ is given by

$$f_X(k)={\left(\frac{f-k+1}{f}\right)}^N-{\left(\frac{f-k}{f}\right)}^N.$$

The expected value of $X$ will be

$$\begin{array}{rl}E(X)& =\sum _{k=1}^{f}k{f}_X(k)\\ & ={\left(\frac{f}{f}\right)}^N+{\left(\frac{f-1}{f}\right)}^N+\cdots +{\left(\frac{1}{f}\right)}^N\\ & =\sum _{k=1}^f{\left(\frac{k}{f}\right)}^N.\end{array}$$

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10

Find the marginal mass functions of the multinomial distribution.

The multinomial distribution is given by

$$f(x_1,x_2,\dots ,x_r)=\{\begin{array}{ll}\frac{n!}{(x_1!)(x_2!)\dots (x_r!)}{p}_1^{x_1}{p}_2^{x_2}\dots p_r^{x_r}& x_i\in {\mathbb{Z}}_{\ge 0},\sum x_i=n\\ 0& \text{otherwise.}\end{array}$$

The marginal mass functions $f_i(x_i)$ are given by

$$\begin{array}{rl}{f}_i(x_i)& =\frac{p_i^{x_i}n!}{x_i!(n-x_i)!}\left(\sum _{\sum x_j=n-x_i,j\ne i}\frac{(n-x_i)!}{{\prod }_{j\ne i}{x}_j!}\prod _{j\ne i}{p}_j^{x_j}\right)\\ & =p_i^{x_i}\left(\genfrac{}{}{0ex}{}{n}{x_i}\right){\left(\sum _{j=1}^r{p}_j-p_i\right)}^{n-x_i}\\ & =\left(\genfrac{}{}{0ex}{}{n}{x_i}\right)p_i^{x_i}(1-p_i{)}^{n-x_i}.\end{array}$$

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11

$f_X(x)=p_1(1-p_1{)}^x$ and $f_Y(x)=p_2(1-p_2{)}^x$, for $x\in {\mathbb{Z}}_{\ge 0}$. The probability generating functions for $X$ and $Y$ are ${\Phi }_X(t)=\frac{p_1}{1-(1-p_1)t}$ and ${\Phi }_Y(t)=\frac{p_2}{1-(1-p_2)t}$. Since $X$ and $Y$ are independent, ${\Phi }_{X+Y}={\Phi }_X{\Phi }_Y$.

$$\begin{array}{rl}{\Phi }_X{\Phi }_Y(t)& =\frac{p_1{p}_2}{(1-(1-p_1)t)(1-(1-p_2)t)}\\ & =p_1{p}_2\sum _{k=0}^{\infty }\left(\sum _{i=0}^k(1-p_1{)}^i(1-p_2{)}^{k-i}\right)t^k\\ & =p_1{p}_2\sum _{k=0}^{\infty }\left((1-p_2{)}^k\left(\frac{1-{\left(\frac{1-p_1}{1-p_2}\right)}^{k+1}}{1-\left(\frac{1-p_1}{1-p_2}\right)}\right)\right)t^k\\ & =\sum _{k=0}^{\infty }\left(\frac{p_1{p}_2}{p_1-p_2}\right)[(1-p_2{)}^{k+1}-(1-p_1{)}^{k+1}]t^k.\end{array}$$

Thus,

$$f_{X+Y}(k)=\left(\frac{p_1{p}_2}{p_1-p_2}\right)[(1-p_2{)}^{k+1}-(1-p_1{)}^{k+1}].$$

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12

Let $f$ be the PMF of $N$. The support of $Ng(N)$ is $\{kg(k){\}}_{k=0}^{\infty }$, and the PMF $\tilde{f}$ of $Ng(N)$ satisfies $\tilde{f}(kg(k))=f(k)$. Thus,

$$\begin{array}{rl}E(Ng(N))& =\sum _{k=0}^{\infty }kg(k)f(k)\\ & =\sum _{k=0}^{\infty }kg(k)\frac{{\lambda }^k}{k!}{e}^{-\lambda }\\ & =\lambda \sum _{k=1}^{\infty }g(k)\frac{{\lambda }^{k-1}}{(k-1)!}{e}^{-\lambda }\\ & =\lambda \sum _{k=0}^{\infty }g(k+1)\frac{{\lambda }^k}{k!}{e}^{-\lambda }\\ & =\lambda E(g(N+1)).\end{array}$$

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13

$$\begin{array}{rl}{\Phi }_X(t)& =\sum _{k=1}^{\infty }\frac{1}{k(k+1)}{t}^k\\ & \\ & =\sum _{k=1}^{\infty }\frac{1}{k}{t}^k-\sum _{k=1}^{\infty }\frac{1}{k+1}{t}^k\\ & \\ & =-\ln (1-t)+\frac{1}{t}\left(-\sum _{k=2}^{\infty }\frac{1}{k}{t}^k\right)\\ & =-\ln (1-t)+\frac{1}{t}(\ln (1-t)+t)\\ & =\ln (1-t)\left(\frac{1}{t}-1\right)+1\end{array}$$

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14

The generating function of $f$ is

$$\begin{array}{rl}\varphi (t)& =\left(\frac{1-p}{1+p}\right)\left(\sum _{m\ge 0}(pt{)}^m+\sum _{m\le 0}{\left(\frac{t}{p}\right)}^m-1\right)\\ & =\left(\frac{1-p}{1+p}\right)\left(\frac{1}{1-pt}+\frac{t}{t-p}-1\right)\end{array}$$ $$\begin{array}{rl}{\varphi }^{\prime }(t)& =\left(\frac{1-p}{1+p}\right)\left(\frac{p}{{\left(1-pt\right)}^2}+\frac{1}{t-p}-\frac{t}{{\left(t-p\right)}^2}\right)\\ {\varphi }^{\prime \prime }(t)& =-\frac{2{\left(p-1\right)}^{2}p\left(p{t}^3-3pt+p^2+1\right)}{{\left(t-p\right)}^3{\left(pt-1\right)}^3}\end{array}$$

Now, $EX={\varphi }^{\prime }(1)=0$. $\text{Var}(X)={\varphi }^{\prime \prime }(1)+{\varphi }^{\prime }(1)-{\varphi }^{\prime }(1{)}^2={\varphi }^{\prime \prime }(1)=\frac{2p}{(p-1{)}^2}$.