Random Variables

Probability Mass Function (pmf)

E (X) = x \in R_{X} \sum x p (x)

E (g (X)) = x \in R_{X} \sum g (x) p (x)

E (a X + b) = a E (X) + E (b)

If $X$ and $Y$ are independent, then

E (f (X) g (Y)) = E (f (X)) E (g (Y))

and if $f$ and $g$ are continuous, then

E (X Y) = E (X) E (Y)

Var (X) = x \in R_{X} \sum (x - μ)^{2} p (x)

Var (a X + b) = a^{2} Var (X)

Var (X) = E (X^{2}) - E (X)^{2}

If $X$ and $Y$ are independent, then

Var(X+Y) = Var (X) + Var (Y)

\frac{E ( X ) - R _{0}}{σ _{X}}

$\int_{- \infty}^{+ \infty} f (x) d x = 1$
$F (a) = \int_{- \infty}^{a} f (x) d x$ , $\frac{d F ( x )}{d x} ∣_{x = a} = f (a)$
$P (X = a) = \int_{a}^{a} f (x) d x = 0$

For two independent random variables $X$ and $Y$

Var (X + Y) = Var (X) + Var (Y)

Var (X + Y) = E (X + Y - (μ_{x} + μ_{y}))^{2} = E ((X - μ_{x}) + (Y - μ_{y}))^{2} = E ((X - μ_{x})^{2} + 2 (X - μ_{x}) (Y - μ_{y}) + (Y - μ_{y})^{2}) = E (X - μ_{x})^{2} + 2 E () ()

X \sim B (n, p)

p (x) = P (X = x) = (x n) p^{x} (1 - p)^{n - x}

E (X) = n p Var (X) = n p (1 - p)

n \to \infty, p \to 0, n p \to λ, B (n, p) \to P o i s (λ)

p (x) = P (X = k) = e^{- λ} \frac{λ ^{k}}{k !}

E (X) = λ Var (X) = λ

Let $X_{1} \sim P (λ_{1})$ and $X_{2} \sim P (λ_{2})$ , if $X_{1}$ and $X_{2}$ are independent, then $X_{1} + X_{2} \sim P (λ_{1} + λ_{2})$

N (μ, σ^{2})

f (x) = \frac{1}{2 π σ ^{2}} e^{- (x - μ)^{2} /2 σ^{2}}

Z = \frac{X - μ}{σ} \sim N (0, 1)

cdf of a standard normal rv is $Φ (x)$ , $f (x) = Φ^{'} (x)$

P (a < x \leq b) = P (\frac{a - μ}{σ} < Z \leq \frac{b - μ}{σ}) = Φ (\frac{b - μ}{σ}) - Φ (\frac{a - μ}{σ})

E (X) = μ V (X) = σ^{2}

f (t) = \frac{Γ ( \frac{v + 1}{2} )}{v π Γ ( \frac{v}{2} )} (1 + \frac{t ^{2}}{v})^{- (v + 1) /2}

f (x; k) = ⎩ ⎨ ⎧ \frac{x ^{\frac{k}{2} - 1} e ^{- \frac{x}{2}}}{2 ^{\frac{k}{2}} Γ ( \frac{k}{2} )}, 0, x > 0 otherwise

X = Z_{1}^{2} + Z_{2}^{2} + \dots + Z_{k}^{2}

$Z_{i} \sim N (0, 1)$
$E (Z_{i}) = 0$
$Var (Z_{i}) = 1 = E (Z_{i}^{2}) - (E (Z_{i})^{0})^{2} = E (Z_{i}^{2})$ , so $E (Z_{i}^{2}) = 1$

E (X) = k