Solutions 3 - Random Variables and Their Distributions

Entry | Notes

PMFs and CDFs

Question 1

$$P(X=k)=p_X(k)=P(\text{person k matches}|\text{first k-1 different})P(\text{first k-1 different})$$ $$p_X(k)=\frac{k-1}{365}\times \frac{365\times 364\times \cdots \times (365-k+2)}{365^{k-1}}=\frac{(k-1)365!}{365^k(365-k+1)!}$$

For $k=2,3,4,\dots ,366$ and $p_X(k)=0$ for $k<2$ or $k>366$.

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Question 2

(a)

$$\begin{array}{rl}P(X=k)& =p_X(k)\\ & =P(\text{trial k success}|\text{first k-1 fail})P(\text{first k-1 fail})\\ & +P(\text{trial k fail}|\text{first k-1 success})P(\text{first k-1 success})\\ & =P(\text{trial k success})P(\text{first k-1 failure})\\ & =\frac{1}{2}\times (\frac{1}{2}{)}^{k-1}\\ & =(\frac{1}{2}{)}^k\end{array}$$

For $k=1,2,3,\dots$

(b)

$$\begin{array}{rl}& P(X=k)\\ & =p_X(k)\\ & =P(\text{trial k success}|\text{first k-1 fail})P(\text{first k-1 fail})\\ & +P(\text{trial k fail}|\text{first k-1 success})P(\text{first k-1 success})\\ & =\frac{1}{2}\times {\left(\frac{1}{2}\right)}^{k-1}+\frac{1}{2}\times {\left(\frac{1}{2}\right)}^{k-1}\\ & =2\times {\left(\frac{1}{2}\right)}^k={\left(\frac{1}{2}\right)}^{k-1}\end{array}$$

For $k\ge 2$ and $p_X(k)=0$ for $k<2$

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Question 3

$$\begin{array}{rl}{F}_Y(y)& =P(Y\le y)\\ & =P(\mu +\sigma X\le y)\\ & =P\left(X\le \frac{y-\mu }{\sigma }\right)\\ & =F_X\left(\frac{y-\mu }{\sigma }\right)\end{array}$$

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Question 4

• It’s an increasing function
• It’s right-continuous:

$$F(a)=\underset{x\to a^{+}}{\lim }F(x)$$

• It converges to $0$ and $1$ at in the limits:

$$\underset{x\to -\infty }{\lim }F(x)=0\text{ and }\underset{x\to \infty }{\lim }F(x)=1$$

The PMF would be:

$$p_X(k)=\frac{1}{n}$$

For $k=1,2,3,\dots ,n$, and otherwise $p_X(k)=0$, since it increases CDF at every integer between $1$ and $n$ by $\frac{1}{n}$.

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Question 5

(a)

• Checking for being non-negative:

$$\text{for }n=0,1,2,\dots ,p(n)={\left(\frac{1}{2}\right)}^{n+1}>0$$

• Checking for the sum:

$$\sum p(n)=\sum _{n=0}^{\infty }{\left(\frac{1}{2}\right)}^{n+1}=\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{2}\right)}^n=\frac{1}{2}\times \frac{1}{1-\frac{1}{2}}=1$$

Since it satisfies both conditions, it’s a PMF.

(b)

$$\begin{array}{rl}{F}_X(k)& =P(X\le k)\\ & =\sum _{n=0}^{\lfloor k\rfloor }{\left(\frac{1}{2}\right)}^{(n+1)}\\ & =\frac{1}{2}\sum _{n=0}^{\lfloor k\rfloor }{\left(\frac{1}{2}\right)}^n\\ & =\frac{1}{2}\times \frac{1-{\left(\frac{1}{2}\right)}^{\lfloor k\rfloor }+1}{1-\frac{1}{2}}\\ & =1-{\left(\frac{1}{2}\right)}^{\lfloor k\rfloor +1}\end{array}$$

For $k\ge 0$

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Question 6

• Checking for being non-negative:

$$P(D=j)={\log }_{10}\left(\frac{j+1}{j}\right)>0\text{ since }\frac{j+1}{j}>1\text{ for }1\le j\le 9$$

• Checking for the sum:

$$\begin{array}{rl}\sum _{j=1}^{9}P(D=j)& =\sum _{j=1}^9{\log }_{10}\left(\frac{j+1}{j}\right)\\ & =\sum _{j=1}^9{\log }_{10}(j+1)-{\log }_{10}(j)\\ & ={\log }_{10}(10)-{\log }_{10}(9)+{\log }_{10}(9)-\cdots -{\log }_{10}(2)+{\log }_{10}(2)-{\log }_{10}(1)\\ & =1\end{array}$$

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Question 7

$$P(X=k)=P(\text{lose level k}|\text{won first k-1 levels})P(\text{won first k-1 levels})$$ $$P(X=k)=(1-p_k)\prod _{i=1}^{k-1}{p}_i$$

for $1\le j\le 6$

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Question 8

$$P(X=k)=P(\text{k is the most valuable}|\text{the other 4 are less valuable})P(\text{the other 4 are less valuable})$$ $$p_{X(k)}=\frac{\left(\genfrac{}{}{0ex}{}{k-1}{4}\right)}{\left(\genfrac{}{}{0ex}{}{100}{5}\right)}$$

For $k=5,6,7,\dots ,100$, and $P(X=k)=0$ otherwise.

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Question 9

(a)

1. Increasing:

$$\frac{dF}{dx}=p\frac{d{F}_1}{dx}+(1-p)\frac{d{F}_2}{dx}$$

Since $\frac{d{F}_1}{dx}\ge 0$ and $\frac{d{F}_2}{dx}\ge 0$ (they are valid CDFs), thus,

$$p\frac{d{F}_1}{dx}\ge 0\text{ and }(1-p)\frac{d{F}_2}{dx}\ge 0$$

Since $p>0$ and $(1-p)>0$, therefore,

$$\frac{dF}{dx}=p\frac{d{F}_1}{dx}+(1-p)\frac{d{F}_2}{dx}\ge 0$$

2. Right-Continuous:

$$\underset{x\to a^{+}}{\lim }F(x)=\underset{x\to a^{+}}{\lim }(p{F}_1(x)+(1-p)F_2(x))=p\underset{x\to a^{+}}{\lim }{F}_1(x)+(1-p)\underset{x\to a^{+}}{\lim }{F}_2(x)$$

Since $F_1$ and $F_2$ are both valid CDFs,

$$\underset{x\to a^{+}}{\lim }F(x)=p{F}_1(a)+(1-p)F_2(a)=F(a)$$

3. Convergence to $0$ and $1$ in the limits: Computing the limit at $-\infty$

$$\underset{x\to -\infty }{\lim }F(x)=\underset{x\to -\infty }{\lim }(p{F}_1(x)+(1-p)F_2(x))=p\underset{x\to -\infty }{\lim }{F}_1(x)+(1-p)\underset{x\to -\infty }{\lim }{F}_2(x)$$

Since $F_1$ and $F_2$ are both valid CDFs,

$$\underset{x\to -\infty }{\lim }F(x)=p\times 0+(1-p)\times 0=0$$

Computing the limit at $\infty$,

$$\underset{x\to +\infty }{\lim }F(x)=\underset{x\to +\infty }{\lim }(p{F}_1(x)+(1-p)F_2(x))=p\underset{x\to +\infty }{\lim }{F}_1(x)+(1-p)\underset{x\to +\infty }{\lim }{F}_2(x)$$

Since $F_1$ and $F_2$ are both valid CDFs,

$$\underset{x\to +\infty }{\lim }F(x)=p\times 1+(1-p)\times 1=1$$

Thus, $F$ is a valid CDF.

(b) Let $X$ be the r.v., Therefore,

$$F_X(x)=P(X\le x)=P(X\le x|\text{Heads})P(\text{Heads})+P(X\le x|\text{Tails})P(\text{Tails})$$

Since $P(X\le x|\text{Heads})=F_1(x)$ and $P(X\le x|\text{Tails})=F_2(x)$, therefore,

$$F_X(x)=p{F}_1(x)+(1-p)F_2(x)$$

Which is equivalent to $F(x)$,

$$F_X(x)=F(x)$$

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Question 10

(a) Assume $p_X(n)$ is such PMF,

$$p_X(n)=P(X=n)=k\frac{1}{n}$$

Therefore, it must,

1. Be non-negative:

$$p_X(n)\ge 0⟹k\frac{1}{n}\ge 0$$

And since $\frac{1}{n}>0$, thus,

$$k\ge 0$$

2. Sum to $1$:

$$\sum _{n=1}^{\infty }{p}_X(n)=k\sum _{n=1}^{\infty }\frac{1}{n}$$

Which equals to,

• $\infty$, for $k>0$
• $-\infty$ for $k<0$
• $0$ for $k=0$

Therefore there is no such discrete distribution.

(b) Assume $p_X(n)$ is such PMF,

$$p_X(n)=P(X=n)=k\frac{1}{n^2}$$

Therefore, it must,

1. Be non-negative:

$$p_X(n)\ge 0⟹k\frac{1}{n^2}\ge 0$$

And since $\frac{1}{n^2}>0$, thus,

$$k\ge 0$$

2. Sum to $1$:

$$\sum _{n=1}^{\infty }{p}_X(n)=k\sum _{n=1}^{\infty }\frac{1}{n^2}=k\frac{{\pi }^2}{6}$$

Therefore, for it to equal to $1$,

$$k\frac{{\pi }^2}{6}=1⟹k=\frac{6}{{\pi }^2}$$

Which is consistent with the above $k\ge 0$ condition.

Therefore there is only one such PMF:

$$p_X(n)=\frac{6}{n^2{\pi }^2}$$

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Question 11

$$F_X(x)=P(X\le x)=P(X<x\cup X=x)=P(X<x)+P(X=x)-P(X<x\cap X=x)$$

Since $P(X<x\cap X=x)=0$ (It’s an impossible event), therefore,

$$F_X(x)=P(X<x)+P(X=x)=G_X(x)+P(X=x)$$

Since it’s a discrete distribution, $P(X=x)=p_X(x)$, thus,

$$F_X(x)=G_X(x)+p_X(x)$$

Therefore,

$$G_X(x)=F_X(x)-p_X(x)$$

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Question 12

(a)

Let:

• $p_X(x)=1$ For $x=2$ and $p_X(x)=0$ otherwise.
• $p_Y(x)=1$ For $x=1$ and $p_Y(x)=0$ otherwise.

Thus,

$$F_X(x)=\{\begin{array}{ll}0& x<2\\ 1& 2\le x\end{array}\text{ and }{F}_Y(x)=\{\begin{array}{ll}0& x<1\\ 1& 1\le x\end{array}$$

Comparing them,

$$\{\begin{array}{ll}x<1& F_X(x)=F_Y(x)=0\\ 1\le x<2& F_X(x)=0<1=F_Y(x)\\ 2\le x& F_X(x)=F_Y(x)=1\end{array}$$

Diagram of PMFs, red for $p_Y(x)$ and blue for $p_X(x)$.

Diagram of CDFs, red for $F_Y(x)$ and blue for $F_X(x)$.

(b)

$$P(X=x)\le P(Y=x)⟹\sum P(X=x)\le \sum P(Y=x)$$

Since ${\lim }_{x\to \infty }{F}_X(x)={\lim }_{x\to \infty }{F}_Y(x)=1$ therefore,

$$\sum P(X=x)=\sum P(Y=x)=1$$

Thus,

$$P(X=x)\le P(Y=x)⟹1\le 1$$

But, since $P(X=x_0)<P(Y=x_0)$ for some $x_0$ (strict inequality), we would need:

$$\sum P(X=x)<\sum P(Y=x)$$

Which gives $1<1$, a contradiction, therefore it’s impossible to have such r.v.s.

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Question 13

By LOTP,

• $P(X=a)={\sum }_{z}P(X=a|Z=z)P(Z=z)$
• $P(Y=a)={\sum }_{z}P(Y=a|Z=z)P(Z=z)$

Since we know $P(X=a|Z=z)=P(Y=a|Z=z)$, for all $a$ and $z$, therefore,

$$\sum _{z}P(X=a|Z=z)P(Z=z)=\sum _{z}P(Y=a|Z=z)P(Z=z)$$

Using the equations above,

$$P(X=a)=P(Y=a)$$

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Question 14

(a)

• $P(X\ge 1)=1-P(X=0)=1-e^{-\lambda }$
• $P(X\ge 2)=1-P(X=0)-P(X=1)=1-e^{-\lambda }(1+\lambda )$

(b)

$$p_X(k)=P(X=k|X\ge 1)=\frac{P(X=k\cap X\ge 1)}{P(X\ge 1)}=\frac{e^{-\lambda }{\lambda }^k/k!}{1-e^{-\lambda }}\text{ for }k\ge 1$$

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Question 15

$$F_X(x)=\{\begin{array}{ll}x<1& 0\\ 1\le x\le n& \frac{\lfloor x\rfloor }{n}\\ x>n& 1\end{array}$$

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Question 16

$$P(X=x|X\in B)=\frac{P(X=x\cap X\in B)}{P(X\in B)}=\frac{\frac{1}{|C|}}{\frac{|B|}{|C|}}=\frac{1}{|B|}⟹(X|X\in B)\sim \text{Dunif}(B)$$