more on CDF

CDF: \(F(x) = P(X \le x), \text{as a function of real x }\)

Average of a discrete r.v. X E: Expected

\[E(X) = \sum_{X} X \times P(X=x)\]

Example:

\[X \sim Bern(p) \\ E(X) = 1 \times P(X=1) + 0 \times P(X=0) = p\]

indicator of r.v

\[X = \begin{cases} 1, \text{if A occurs} \\ 0, \text{otherwise} \end{cases} \\ \text{then } \ E(X)=P(A)\]

Expectation of Binomial

\[X \sim Bin(n,p) \\ \begin{align} E(X) & = \sum_{k=0}^{n} k \binom{n}{k} p^k q^{n-k} \\ & = \sum_{k=0}^{n} n \binom{n-1}{k-1} p^k q^{n-k} ,\text{hint:} k \binom{n}{k} = n \binom{n-1}{k-1} \\ & = n p \sum_{k=1}^{n} \binom{n-1}{k-1} p^{k-1} q^{n-k} ,\text{hint: k=0 is zero, and let j=k-1} \\ & = n p \sum_{j=0}^{n-1} \binom{n-1}{j} p^{j} q^{n-1-j} \\ & = n p ( p + q )^{n-1} \\ & = n p \end{align}\]

Can also solved by Linearity

\[\text{Linearity:} \\ E(X+Y) = E(X) + E(Y), \text{even if X,Y are dependent} \\ E(C X) = C E(X), \text{C is a constant} \\\]

Redo by Linearity: sum of n iid Bern(p) is \(n \times p\) \(\text{Since} \ X = X_1 + X_2 + X_3 + ... + X_n, X_j \sim Bern(p)\)

Expected Value of Hypergeometric

5 cards from a deck, X = (# aces)

\[\text{Let} \ X_j \text{be indicator of jth card being an ace,} \ 1 \le j \le 5 \\ \begin{align} E(X) & = E(X_1 + X_2 + ... + X_5) , \text{by indicator random variable} \\ & = E(X_1) + E(X_2) + ... + E(X_5), \text{by linearity} \\ & = 5 \times E(X_1), \text{by symmetry} \\ & = 5 \times P( \text{1st card is ace} ), \text{by fundamental bridge} \\ & = \frac{5}{13} \end{align}\]

Poisson Paradigm (Pois Approximation)

\[\text{Events} \ A_1, A_2, ..., A_n. P(A_j) = p_j, \\ n \text{ is large,} \ p_j \text{'s small}\]

Events are independent or “weakly independent”, then # of \(A_j\)’s that occur is approx. \(Pois(\lambda), \lambda = \sum_{j=1}^{n} p_j\)

\[Bin(n, p), let \ n\to\infty, p \to 0, \lambda = np\ \text{is held constant.} \\ \text{Find what happends to } P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}, k \ \text{fixed}\] \[\begin{align} P(X=k) & = \frac{n(n-1)...(n-k+1) \lambda^k }{k! \ n^k} \times (1-\frac{\lambda}{n})^n \times (1-\frac{\lambda}{n})^{-k} \\ & = \frac{\lambda^k}{k!} \times ( \frac{n}{n} \frac{n-1}{n} ... \frac{n-k+1}{n} ) \times e^{-\lambda} \times 1 \\ & = \frac{\lambda^k}{k!} \times e^{-\lambda} \\ & = Pois PMF \end{align}\]

Hints: \(p = \frac{\lambda}{n}\) \(\binom{n}{k} = \frac{n(n-1)...(n-k+1)}{k!}\) \((1 + \frac{x}{n})^n \to e^x \ as \ n \to \infty\) \((1-\frac{\lambda}{n})^{-k} \to 1\)

Birthday problem

Having n people, find approximation Prob. that there are 3 people with the same birthday.

\(\binom{n}{3}\) triplets of people. indicator of r.v. for each, For i,j,k, i < j < k E(# triple matches) = \(\binom{n}{3} \frac{1}{365^2}\)

X = # triple matches. Approx \(Pois(\lambda), \lambda = \binom{n}{3} \frac{1}{365^2}\)

\[P(X \ge 1) = 1 - P(X=0) \approx 1 - e^{-\lambda} \frac{\lambda^0}{0!} = 1 - e^{-\lambda}\]