# Interview

Welcome to your variance interview! In a quantitative finance interview, one of the most important things is to explain your thoughts. A good thought process is worth more than a correct answer with no ability to articulate your reasoning. This quiz is no different, as we’ll be breaking down a challenging problem into its logical steps.

## Interview

### Variance

# Interview

Here’s the problem we’ll be tackling:

You are playing a game in which you will be paid $1 for each pair of consecutive heads you flip out of 10 coins. For example, if you flip THHHTTHHTT, you will receive$3. What is the variance of your payout?

## Interview

### Variance

# Interview

It makes sense to set indicator variables for the potential “successful” events. How many of these indicator variables should you have?

## Interview

### Variance

# Interview

If we let $$X_k$$ be an indicator variable on the event that flips $$k$$ and $$k+1$$ are both heads, what is $$\text{var}(X_k)?$$

## Interview

### Variance

# Interview

In the previous question, we found that $$\text{var}(X_k) = \frac{3}{16}.$$ Since the total payout is $$\sum_{k=1}^{9}X_k,$$ is it true that the variance of the payout is $$9 \cdot \frac{3}{16} = \frac{27}{16}?$$

## Interview

### Variance

# Interview

Right, we can’t add the variances of the $$X_k,$$ since at least some of them are dependent. Using the identity $$\text{var}(X) = E(X^2) - \left(E(X)\right)^2,$$ we can write the variance of the payout as $E\left(\sum_{k=1}^{9}X_k^2 + \sum_{k \ne j} X_kX_j \right) - \left(E\left(\sum_{k=1}^9 X_k\right)\right)^2.$

We can compute each part of this variance separately. For starters, what is the expected value of the game, $E\left(\sum_{k=1}^9 X_k\right)?$

## Interview

### Variance

# Interview

We’ve got the variance down to $E\left(\sum_{k=1}^{9}X_k^2 + \sum_{k \ne j} X_kX_j \right) - 2.25^2.$ What is $E\left(\sum_{k=1}^{9}X_k^2\right)?$

## Interview

### Variance

# Interview

We’ve got the variance down to $2.25 + E\left(\sum_{k \ne j} X_kX_j \right) - 2.25^2.$ What is $E\left(\sum_{k \ne j} X_kX_j \right)?$

Hint: There are a total of $$9 \cdot 8$$ pairs $$(X_k,X_j)$$ in the sum. Some of these are pairs of independent random variables, while others are of dependent random variables.

## Interview

### Variance

# Interview

What is the variance of the payout?

## Interview

### Variance

×