# Interesting Statsitcal Problems in Interviews

### Problem 1

If we have $n$ dice draws with a dice that has $k$ sides. Let $X_i$ to be the random variables that describes the number of times that the dice faces $i$. What is the correlation between $x_i$ and $x_j$ for $i\neq j$.

We first notice, $X_i$ are identitically distributed. Thus, the correlation between $X_i$ and $X_j$ does depend on $i$ and $j$ and they they have the same variance. Thus, we only need to calculate covariance first. We then notice

$X_1+X_2 + \dots + X_k = n$

Next we compute

\begin{aligned} &\text{Cov}[X_i, X_1+X_2+\dots + X_k] = \text{Cov}[X_i, n] =0 \\ &=\text{Cov}(X_i, X_i) + (k-1)\text{Cov}(x_i,x_j)\\ \end{aligned}

Thus, we find

$(k-1)\text{Cov}(x_i,x_j) = - \text{Var}(X_i)$

Thus,

$\text{Cov}(x_i,x_j) = - \frac{\text{Var}(x_i)}{k-1}$

Therefore, the correlation becomes

$\text{Corr}(X_i,X_j) = \frac{\text{Cov}(X_i,X_j)}{\text{Var}(X_i)} = - \frac{1}{k-1}$

### Problem 2

With a set of 52 cards, we draw without replacement from the card. What is the expected number of draws to see the first Ace?

Counting down from the top of the deck of cards. Let $X_1$ be the number of cards before the first Ace. Let $X_2$ be the number of cards between the first and the second Ace. Let $X_3$ be the number of cards between the second and third Ace. Let $X_4$ be the number of cards between the third Ace and the fourth Ace. Finally, let $X_5$ be the number of cards between the fourth Ace and the bottom of the deck. We find

$X_1+X_2+X_3+X_4+X_5 = 52-4 = 48$

We also find the distribution of $X_i$’s are the same and thus,

$5E[X_1] = 48\quad \Rightarrow \quad E[X_1] = \frac{48}{5}$

### Problem 3

A 2D random walk starts from $(1,1)$. If it will stop once it hits the $y$-axis. What is the probability that it will stop at the negative part of the $y$-axis?

Notice the symmetry of the problem. In 2D random walk, the problem is symmetric. At a given time, it will be uniformly distributed in the direction respect to the starting point. Given it hits $y$-axis, it will be uniformly distributed around within the angle ($0$,$\pi$) with respect to the starting point. However, the probability of going to the negative part is the last $\pi/4$ angle. Thus, the probability of landing in the negative part of $y$-axis is $1/4$.

### Problem 4

Give an example that two random variables are uncorrelated but dependent

Let $X = \sin(\tau)$ and $Y = \cos(\tau)$ and $\tau\sim U(0,2\pi)$. This is uncorrelate becuase, if we plot $(X,Y)$, it relys on the unit circle. If we run a linear regression on that, the line will have zero slope. Since correlation between $Y$ and $X$ is the $R^2$ score and it is zero in this case. They are dependent because, once we know $X$, $Y$ can only have at most two values.

### Problem 5

How to uniformly generate points on a disc with radiu $R$?

If we label points by $(r,\theta)$, the probability density function of $(r,\theta)$ is

$p(r,\theta)drd\theta = \frac{rd\theta dr }{\pi R^2} = \frac{d\theta}{2\pi} \cdot \frac{d r^2}{R^2} = p(\theta) p(r^2)$

Thus, we uniformly generate $r^2$ in the interval $[0,R^2]$ and uniform generate $\theta$ from $[0,2\pi)$.

Categories:

Updated: