Peer Sheet 3

P3.1 Distributions

a)

$X$ - the number of bits flipped $X\sim \text{Bin}(n,p)$ The probability that exactly $k$ bits are flipped is: $\mathrm{Pr}[X=k]=\left(\begin{array}{c}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}$

b)

We define one round as “Alice tosses once and if not succeed then Bob tosses once” A round is not succeed if both of them fail, which has a probability of $\frac{1}{2}(1-p)$

Let $A_k$ be the event that Alice wins at round $k$ $\mathrm{Pr}[A_k]=(\frac{1}{2}(1-p){)}^{k-1}\cdot \frac{1}{2}$ The total probability that Alice wins ($A$) is:

$$\mathrm{Pr}[A]=\sum _{k=1}^{\infty }\mathrm{Pr}[A_k]=\sum _{k=1}^{\infty }(\frac{1}{2}(1-p){)}^{k-1}\cdot \frac{1}{2}=\frac{\frac{1}{2}}{1-\frac{1}{2}(1-p)}=\frac{1}{1+p}$$

In order to make the game fair we must have $\mathrm{Pr}[A]=\frac{1}{2}$ This means $\frac{1}{1+p}=\frac{1}{2}⟹p=1$ However, the biased coin has $p<1$. This leads to a contradiction, which means there is no valid $p$ that makes the game fair.

c)

X

We first consider a single plate: Let random variable $X_i$ be the number of caught throws before the plate $i$ breaks. $X_i$ follows geometric distribution, therefore: $\mathbb{E}[X_i]=\frac{1-p}{p}=1$ and $\text{Var}[X_i]=\frac{1-p}{p^2}=2$

By using the linearity of expected value we have

$$\mathbb{E}[X]=\mathbb{E}\left[\sum _{i=1}^n{X}_i\right]=\sum _{i=1}^n\mathbb{E}[X_i]=\sum _{i=1}^{n}1=n\cdot 1=n$$

and by using Hint ($X_i$ are independent variables)

$$\text{Var}[X]=Var\left[\sum _{i=1}^n{X}_i\right]=\sum _{i=1}^n\text{Var}[X_i]=\sum _{i=1}^{n}2=2n$$

Y

Since all plates are going to be thrown until broken. Therefore, the number of throws that are not caught is always $n$ $\mathbb{E}[Y]=n$ $\text{Var}[Y]=\mathbb{E}[(Y-\mathbb{E}[Y]){]}^2=\mathbb{E}[(0-0{)}^2]=0$

X+Y

By using linearity of expected value:

$$\mathbb{E}[X+Y]=\mathbb{E}[X]+\mathbb{E}[Y]=n+n=2n$$

Since $Y$ is a constant and adding a constant to a random variable does not change its variance:

$$\text{Var}[X+Y]=\text{Var}[X+n]=\text{Var}[X]=2n$$