Here are my original hand-written notes and more on this topic.

A *Bernoulli($p$)* random variable is a random variable that takes on a 1 with probability $p$ and a 0 with probability $1-p$.

If $X$ is Bernoulli($p$), then \[ \E[X] = p\cdot 1 + (1-p)0 = p \enspace . \] (These are named after Jacob Bernoulli, a mathematician who lived in the 16-1700s.)

A *binomial($p,n$)* random variable is the sum of $n$ independent Bernoulli($p$) random variables.
\[
B = X_1+X_2+\cdots + X_n
\]
If $B$ is a binomial$(p,n)$, then,
\[
\E[B] = \E\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n \E[X_i] = pn \enspace .
\]

# The Distribution of Binomial Random variables

A binomial random variable takes on values in $\{0,\ldots,n\}$, but the distribution is clearly not uniform. In fact, \[ \Pr\{B=i\} = \binom{n}{i}p^i(1-p)^{n-i} \enspace , \] but this is kind of awkward to work with. (Try computing $\E[B]$ without using linearity of expectation and you'll understand.)

Chernoff's Bounds give us estimates on the probability that a binomial$(n,p)$ is far from its expected value. For any $0 < \epsilon < 1$, \[ \Pr\{B \ge (1+\epsilon)pn\} \le e^{-\epsilon^2pn/3} \enspace , \] and \[ \Pr\{B \le (1-\epsilon)pn\} \le e^{-\epsilon^2pn/2} \enspace . \] Notice that these probability get exponentially small in $n$.

# Application: Estimating Sample Sizes

Suppose we have $n$ elements and we sample each element indpendently with probabilty $1/2$. If $B$ is the number of samples, then $B$ is a binomial$(1/2,n)$ random variable, so $\E[B] = n/2$.

It might be a problem if we have too many samples (maybe we don't have enough memory). Chernoff covers this: \[ \Pr\{B \ge (1+\epsilon) n/2\} \le e^{-\epsilon^2 n/6} \]

It might be a problem if we have too few samples (maybe some calculation will be incorrect). Chernoff covers this too: \[ \Pr\{B \le (1-\epsilon) n/2\} \le e^{-\epsilon^2 n/4} \]

So the sample size is very close to $n/2$ with probability very close to 1.

# Application: Monte-Carlo Integral Evaluation

Suppose we have some weird function $f(x)$. We know $1/2\le f(x)\le 1$ for all $x\in[0,1]$. We know how to evaluate $f(x)$, but it's so weird we don't know how to integrate it.

Evaluating the integral $\int_0^1 f(x)\, dx$ is the same as asking for the area under the curve defined by $f(x)$. Here's an algorithm:

- $C\gets 0$
- Pick $(x,y)$ uniformly at random in the unit square $[0,1]^2$.
- If $f(x) > y$ then add 1 to $C$
- Repeat steps 1 and 2 $k=c\ln n$ times
- Output $C/k$

Observe that $C$ is a binomial$(p,n)$ random variable where $p=\int_0^1 f(x)\, dx$ is exactly the value we are trying to compute. And, by assumption, $p>1/2$. So, by Chernoff's bounds: \[ \Pr\{C \ge (1+\epsilon)pk\} \le e^{-\epsilon^2 pk/3} \le e^{-\epsilon^2 c\ln n/6} = n^{-\epsilon^2 c/6} \] and \[ \Pr\{C \le (1-\epsilon)pk\} \le e^{-\epsilon^2 pk/2} \le e^{-\epsilon^2 c\ln n/4} = n^{-\epsilon^2 c/4} \enspace . \]

# Application: Randomized Binary Search

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