# Stephen Malina

## This is my blog. There are many others like it but this one is mine.

# All of Statistics - Chapter 3

# Selected Exercises

**1.** Suppose we play a game where we start with $ c $ dollars. On each play of the game you either double or halve your money, with equal probability. What is your expected fortune after $ n $ trials?

I’m pretty sure we can use a trick that I’ve seen a lot in machine learning to solve this using plain old expectations. If I’m right, we let $ X \sim \text{Binomial}(n, \frac{1}{2}) $ and can create a new random variable, $ Z $, where

$$ Z = c \cdot 2^{X} \left(\frac{1}{2}\right)^{1-X}. $$

Taking the expectation when $ n = 1 $ is easy. We just get
$$
\begin{align}
\mathbb{E} Z &= c \cdot \mathbb{E}\left[2^{X} \left(\frac{1}{2}\right)^{1-X} \right] \\\

&= c \cdot \left(2 \cdot 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} \cdot \frac{1}{2} \right) \\\

&= \frac{5}{4} c .
\end{align}
$$

For $ n > 1 $, I initially thought this was going to be really hard. But thinking about it a bit more, it seems like the fact that a binomial just equals the sum of many Bernoulli random variables makes this easy too. Formally, since $ X = X_1 + \cdots + X_n $ for some $ n $ i.i.d. Bernoulli random variables, $$ \mathbb{E} Z = c \cdot \mathbb{E}\left[2^{X} \left(\frac{1}{2}\right)^{1-X} \right] = c \cdot \prod_{i=1}^n\mathbb{E}\left[2^{X_i} \left(\frac{1}{2}\right)^{1-X_i} \right] = \left(5 / 4\right)^n c. $$

**2.** Show that $ \mathbb{V}(X) = 0 $ iff $ P(X = c) = 1 $ for some $ c $.

The reverse direction is easy. We just note that $ P(X = c) = 1 \implies \mathbb{E}[X] = c \land \mathbb{E}[X^2] = c^2 $, so $$ \mathbb{V}(X) = \mathbb{E}\left[X^2\right] - \mathbb{E}[X]^2 = c^2 - c^2 = 0. $$

For the forward direction, we have to work things about a bit more explicitly. The formula for $ \mathbb{V}(X) $ tells us
$$
\begin{align}
\mathbb{E}\left[\left(X-\mu_X\right)^2\right] &= 0 \\\

\sum_{x_i \in X} (x_i-\mu_X)^2 \Pr(X=x) &= 0.
\end{align}
$$

Because of the probability axioms and the squared term, each term in the sum is always non-negative and the probabilities have to sum to 1. So, the only way this sum can equal 0 is if $ \Pr(X = \mu_X) = 1 $ for an arbitrary choice of $ \mu_X $. The same thing holds for a continuous distribution, since its density can never be negative. One way to see this is to note that the density is the derivative of the CDF, and so it’s always non-negative.