Let \(X = \mu + \sigma Z\)

\(\mu \in \mathbb{R}\) (mean, location)

\(\sigma > 0\) (SD, scale)

Then we say \(X \sim N(\mu, \sigma^2)\)

\[E(X) = \mu\] \[Var(\mu + \sigma Z)= \sigma^2 Var(Z) = \sigma^2\]

Standardization: \(Z = \frac{X-\mu}{\sigma}\)

\[Var(X) = E \left( (X-EX)^2 \right) = EX^2 - (EX)^2 \\ Var(X + c) = Var(X) \\ Var(X \times c) = c^2 \times Var(X) \\\]

Find PDF of \(N(\mu, \sigma^2)\)

\[CDF: P(X \le x) = P(\frac{X-\mu}{\sigma} \le \frac{x-\mu}{\sigma}) \\ = \Phi(\frac{X-\mu}{\sigma}) \\ \Rightarrow PDF = \frac{1}{\sigma\sqrt{2\pi}} e^{-(\frac{X-\mu}{\sigma})^2/2} \\ -X = -\mu + \sigma(-Z) \sim N(-\mu, \sigma^2)\]

later will show:

\[\begin{align} if \ X_j \sim N(\mu,\sigma^2) indep., & X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2) \\ & X_1 - X_2 \sim N(\mu_1 - \mu_2, \sigma_1^2 + \sigma_2^2) \\ \end{align}\]

68-95-99.7% Rule

\[X \sim N(\mu, \sigma^2) \\ P( |X-\mu| \le \sigma ) \approx 0.68 \\ P( |X-\mu| \le 2 \ \sigma ) \approx 0.95 \\ P( |X-\mu| \le 3 \ \sigma ) \approx 0.997\]

LOTUS:

\[X: r.v. \in \{ 0, 1, 2, ... \}\] \[\begin{array}{cccccc} PMF: & P_0 & P_1 & P_2 & P_3 & ... \\ X: & 0 & 1 & 2 & 3 & ... \\ X^2: & 0^2 & 1^2 & 2^2 & 3^2 & ... \end{array} \\ E(X) = \sum_x x P(X=x) \\ E(X^2) = \sum_x x^2 P(X=x) \\\] \[X \sim Pois(\lambda) \\ \begin{align} E(X^2) & = \sum_{k=0}^{\infty} k^2 \frac{e^{-\lambda} \ \lambda^k}{k!} \\ & = e^{-\lambda} \times \lambda \ e^\lambda(\lambda + 1) \\ & = \lambda \ (\lambda + 1) \\ \end{align}\] \[Var(X) = \lambda \ (\lambda + 1) - \lambda^2 = \lambda\]

Taylor series \(\sum_{k=0}^{\infty} \frac{\lambda^k}{k!} = e^\lambda\) take derivatives of both sides multiply be lambda take derivatives of both sides again

\[\begin{align} \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} & = e^\lambda \\ \sum_{k=1}^{\infty} \frac{k \ \lambda^{k-1}}{k!} & = e^\lambda \\ \lambda \sum_{k=1}^{\infty} \frac{k \ \lambda^{k-1}}{k!} & = \lambda \ e^\lambda \\ \sum_{k=1}^{\infty} \frac{k \ \lambda^k}{k!} & = \lambda \ e^\lambda \\ \sum_{k=1}^{\infty} \frac{k^2 \ \lambda^{k-1}}{k!} & = \lambda \ e^\lambda + e^\lambda = e^\lambda(\lambda + 1) \\ \sum_{k=1}^{\infty} \frac{k^2 \ \lambda^k}{k!} & = \lambda \ e^\lambda(\lambda + 1) \end{align}\]

X ~ Bin(n,p), Find Var(X)

if X, Y indep., Var(X+Y) = Var(X) + Var(Y) Binomial is sum of n indep. Bern(p)

\[I_j \sim i.i.d. of Bern(p)\]

\(I_1 \times I_2\) : indicator of success on both trials 1st, 2nd.

\[\begin{align} X & = I_1 + ... + I_n \\ X^2 & = I_1^2 + ... + I_n^2 + 2 I_1 I_2 + 2 I_1 I_3 + ... + 2 I_{n-1} I_n \\ E(X^2) & = n E(I_1^2) + 2 \binom{n}{2} E(I_1 I_2) \\ & = n p + n ( n - 1 ) p^2 \\ & = n p + n^2 p^2 - n p^2 \\ Var(X) & = E(X^2) - (EX)^2 = np - n p^2 = np(1-p) = npq \end{align}\]

Prove LOTUS for discrete sample space.

Show \(E( g(x) ) = \sum_x g(x) P(X=x)\)

\[\sum_x g(x) P(X=x) \\ = \sum_{s \in S} g \left( X(s) \right) \times P( \{s\} ) \\ \sum_x \sum_{s: X(s)=x} g \left( X(s) \right) \times P( \{s\} ) \\ = \sum_x g(x) \sum_{s: X(s)=x} P( \{s\} ) \\ = \sum_x g(x) P(X=x)\]

grouped case / ungrouped case