Probability
\[P(A \cup B) = P(A) + P(B) - P(A\cap B)\]
\[P(A\cap B) = P(A)P(B|A) \]
\[P(A|B) = \frac{P(A\cap B}{P(B)}\]
\[P(A|B) = \frac{P(A)P(B | A)}{P(A)P(B|A)+P(A_c)P(B|A_c)}\]
Binomial distribution
\[\left(\begin{array}{c}n\\ y\end{array}\right)p^{y}(1-p)^{n-y}\]
\[E(Y) = np\]
\[V(Y) = np(1-p)\]
\[My(t) = ((1-p)+pe^t)^{n}\]
Hypergeometric distribution
\[\frac{\left(\begin{array}{c}r\\ y\end{array}\right) \times \left(\begin{array}{c}N-r\\ n-y\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}\]
\[E(Y) = \frac{nr}{N}\]
\[V(Y) = n\left(\frac{r}{N}\right)\left(\frac{N-r}{N}\right)\left(\frac{N-n}{N-1}\right)\]
V(Y) & E(Y)
\[E(Y) = \frac{\sum x}{n}\]
\[V(Y) = { \frac{\sum (x_i - mean)^{2}}{n-1} } = {
\frac{ \sum x^{2} - (\sum x)^{2}/n }{n-1}}\]
If Y = a+bX then:
\[E(Y) = a+bE(X)\]
\[V(Y) = b^2E(X)\]
Poisson distribution
\[p(y) = \frac{\lambda^{y}e^{\lambda}}{y!}\]
\[E(Y) = \lambda\]
\[V(Y) = \lambda\]
\[My(t) = e^{\lambda(e^t-1)}\]
Poisson distribution
\[p(y) = \frac{\lambda^{y}e^{\lambda}}{y!}\]
\[E(Y) = \lambda\]
\[V(Y) = \lambda\]
\[My(t) = e^{\lambda(e^t-1)}\]
Z-test
One sample Z-test
\[Z = \frac{\bar{x}-\mu}{s/\sqrt{n}}\]
Two sample Z-test
\[Z = \frac{\bar{x_1}-\bar{x_2}}{\sqrt{\frac{\sigma_1^{2}}{n_1}+
\frac{\sigma_2^{2}}{n2}}}\]
One Sample proportion test
\[Z = \frac{p_1-p_2}{\sqrt{\frac{p_1+p_2}{n}}}\]
Two sample proportion test
\[Z = \frac{(p_1-p_2)}{\sqrt{P(1-P)\frac{1}{n_1}}}\]
where
\[P = \frac{n_1\times p_1+n_2\times p_2}{n_1+n_2}\]
Geometric Distribution
\[p\times(1-p)^{y-1}\]
\[E(Y) = \frac{1}{p}\]
\[V(Y) = \frac{1-p}{p^2}\]
\[My(t) = \frac{pe^t}{1-(1-p)e^t}\]
Exponential Distribution
\[f(y) = \frac{1}{\beta}e^{-y/\beta}\]
\[F(y) = \int \frac{1}{\beta}e^{\frac{-X}{\beta}} dX\]
\[P(X>x) =e^{-\lambda x}\]
\[P( X< x) = 1-e^{-\lambda x}\]
\[E(Y) = \beta\]
\[V(Y) = \beta^2\]
\[My(t) = (1-\beta t)^{-1}\]
T-test
Two sample t-test
\[t = \frac{\bar{x}_1 - \bar{x}_2 }{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)} }\]
where
\[s_p^2 = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}\]
One sample t-test
\[t = \frac{\bar{x}-\mu}{s/\sqrt{n}}\]
Normal distribution
\[f(y) = \frac{1}{ \sigma \sqrt{2\pi}}e^{-0.5(\frac{x-\mu}{\sigma})^2}\]
\[E(Y) = \mu\]
\[V(Y) = \sigma^2\]
\[My(t) = e^{\mu t+\frac{t^2\sigma^2}{2}}\]
Uniform distribution
\[f(y) = \frac{1}{b-a}\]
\[F(y) = \int_{a}^{b} \frac{1}{b-a}\]
\[E(Y) = \frac{a+b}{2}\]
\[V(Y) = \frac{(b-a)^2}{12}\]
\[My(t) = \frac{e^{tb}-e^{ta}}{t(b-a)}\]