st2334: add until week 5
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@@ -2,15 +2,15 @@
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#set text(size: 9pt)
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#set text(size: 9pt)
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#set list(spacing: 1.2em)
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#set list(spacing: 1.2em)
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- *Mututally Exclusive* - $A inter B = emptyset$
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- *Mututally Exclusive*: $A inter B = emptyset$
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- *Union* - $A union B = { x : x in A or x in B }$
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- *Union*: $A union B = { x : x in A or x in B }$
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- *Intersection* - $A inter B = { x : x in A and x in B }$
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- *Intersection*: $A inter B = { x : x in A and x in B }$
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- *Complement* - $A' = { x : x in S and x in.not A }$
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- *Complement*: $A' = { x : x in S and x in.not A }$
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- $(A inter B)' = (A' union B')$
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- $(A inter B)' = (A' union B')$
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- *Multiplication* - R experiments performed sequentially. Then $n_i dot ... dot n_r$ possible outcomes for $r$ experiments
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- *Multiplication*: R experiments performed sequentially. Then $n_i dot ... dot n_r$ possible outcomes for $r$ experiments
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- *Addition* - $e$ can be performed $k$ ways, and $k$ ways do not overlap : total ways: $n_1 + ... + n_k$
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- *Addition*: $e$ can be performed $k$ ways, and $k$ ways do not overlap : total ways: $n_1 + ... + n_k$
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- *Permutation* - Arrangement of $r$ objects out of $n$, _ordered_. $P^n_r = n!/(n-r)!, P^n_n = n!$
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- *Permutation*: Arrangement of $r$ objects out of $n$, _ordered_. $P^n_r = n!/(n-r)!, P^n_n = n!$
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- *Combination* - Selection of $r$ objects out of $n$, _unordered_ $vec(n, r) = n!/(r!(n-r)!), vec(n, r) times P^r_r = P^n_r$
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- *Combination*: Selection of $r$ objects out of $n$, _unordered_ $vec(n, r) = n!/(r!(n-r)!), vec(n, r) times P^r_r = P^n_r$
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== Probability
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== Probability
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- Axioms:
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- Axioms:
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@@ -30,14 +30,14 @@
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- $P(A inter B) = P(B|A)P(A)$
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- $P(A inter B) = P(B|A)P(A)$
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- $P(A|B) = (P(A)P(B|A)) / P(B)$
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- $P(A|B) = (P(A)P(B|A)) / P(B)$
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- $P(A inter B inter C) = P(A)P(B|A)P(C|B inter A)$
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- $P(A inter B inter C) = P(A)P(B|A)P(C|B inter A)$
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- *Independent* - $P(A inter B) = P(A)P(B), A perp B$
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- *Independent*: $P(A inter B) = P(A)P(B), A perp B$
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- If $P(A) != 0, A perp B arrow.l.r P(B|A) = P(B)$ (Knowledge of $A$ does not change $B$)
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- If $P(A) != 0, A perp B arrow.l.r P(B|A) = P(B)$ (Knowledge of $A$ does not change $B$)
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- *Independence vs Mutually exclusive* -
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- *Independence vs Mutually exclusive*
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- $P(A) > 0 and P(B) > 0, A perp B arrow.double.r "not mutually exclusive"$
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- $P(A) > 0 and P(B) > 0, A perp B arrow.double.r "not mutually exclusive"$
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- Partition - $A_i...A_n$ is mutually exclusive and $union.big^n_i=1 A_i = S, A_i...A_n$ is partition of S
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- Partition: $A_i...A_n$ is mutually exclusive and $union.big^n_i=1 A_i = S, A_i...A_n$ is partition of S
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- $P(B) = sum^n_(i=1) P(B inter A_i) = sum^n_(i=1) P(A_i)P(B|A_i)$
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- $P(B) = sum^n_(i=1) P(B inter A_i) = sum^n_(i=1) P(A_i)P(B|A_i)$
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- $n = 2, P(B) = P(A)P(B|A) + P(A')P(B|A')$
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- $n = 2, P(B) = P(A)P(B|A) + P(A')P(B|A')$
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- *Bayes Theorem* - $P(A_k|B) = (P(A_k)P(B|A_k)) / (sum^n_(i=1)P(A_i)P(B|A_i))$
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- *Bayes Theorem*: $P(A_k|B) = (P(A_k)P(B|A_k)) / (sum^n_(i=1)P(A_i)P(B|A_i))$
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- $n = 2, P(A|B) = (P(A)P(B|A)) / (P(A)P(B|A) + P(A')P(B|A'))$
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- $n = 2, P(A|B) = (P(A)P(B|A)) / (P(A)P(B|A) + P(A')P(B|A'))$
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== Random Variables
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== Random Variables
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@@ -56,9 +56,9 @@
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+ $a <= b, P(a <= X <= b) = integral^b_a f(x) dif x$
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+ $a <= b, P(a <= X <= b) = integral^b_a f(x) dif x$
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- To validate, check (1) and (2)
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- To validate, check (1) and (2)
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- CDF (Discrete) - $F(X) = P(X <=x)$
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- CDF (Discrete): $F(X) = P(X <=x)$
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- $P(a<=X<=b) = P(X<=b) - P(X<a) = F(b) - F(a-), a-$ (is largest value in $R_X$ smaller than $a$)
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- $P(a<=X<=b) = P(X<=b) - P(X<a) = F(b) - F(a-), a-$ (is largest value in $R_X$ smaller than $a$)
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- CDF(Continuous) - $F(X) = integral^x_(-infinity)f(t)dif t$
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- CDF(Continuous): $F(X) = integral^x_(-infinity)f(t)dif t$
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- $f(x) = dif/(dif t) F(x)$
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- $f(x) = dif/(dif t) F(x)$
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- $P(a<=X<=b) = F(b) - F(a)$
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- $P(a<=X<=b) = F(b) - F(a)$
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- $F(x)$ is non-decreasing.
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- $F(x)$ is non-decreasing.
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@@ -76,10 +76,10 @@
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+ $E(X + Y) = E(X) + E(Y)$
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+ $E(X + Y) = E(X) + E(Y)$
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+ Let $g(dot)$ be arbitrary function.
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+ Let $g(dot)$ be arbitrary function.
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- $ E[g(X)] = sum g(x)f(x) \ "or"\ E[g(X)] = integral_R_X g(x)f(x) $
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- $ E[g(X)] = sum g(x)f(x) \ "or"\ E[g(X)] = integral_R_X g(x)f(x) $
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- example - $E(X^2) = sum x^2f(x)$
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- example: $E(X^2) = sum x^2f(x)$
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- Variance - $ sigma^2_x = V(X) = E(X - mu)^2 = E(X^2) - E(X)^2 $
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- Variance: $ sigma^2_x = V(X) = E(X - mu)^2 = E(X^2) - E(X)^2 $
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- Discrete - $V(X) = sum (x-mu_x)^2 f(x)$
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- Discrete: $V(X) = sum (x-mu_x)^2 f(x)$
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- Continuous - $V(X) = integral^(infinity)_(-infinity) (x-mu_x)^2 f(x)$
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- Continuous: $V(X) = integral^(infinity)_(-infinity) (x-mu_x)^2 f(x)$
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- *Properties*
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- *Properties*
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+ $V(a X + b) = a^2V(X)$
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+ $V(a X + b) = a^2V(X)$
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+ $V(X) = E(X^2) - E(X)^2$
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+ $V(X) = E(X^2) - E(X)^2$
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@@ -99,10 +99,68 @@
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+ $integral^infinity_(-infinity)integral^infinity_(-infinity)f_(X,Y)(x, y) dif x dif y= 1$
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+ $integral^infinity_(-infinity)integral^infinity_(-infinity)f_(X,Y)(x, y) dif x dif y= 1$
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=== Marginal Probability Distribution
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=== Marginal Probability Distribution
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- Discrete - $f_X (x) = sum_y f_(X,Y)(x,y)$
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- Discrete: $f_X (x) = sum_y f_(X,Y)(x,y)$
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- Continuous - $f_X (x) =integral^infinity_(-infinity) f_(X,Y)(x,y) dif y$
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- Continuous: $f_X (x) =integral^infinity_(-infinity) f_(X,Y)(x,y) dif y$
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- Conditional Distribution - $ f_(Y|X) (y|x) = (f_(X,Y)(x,y)) / (f_X (x)) $
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- Conditional Distribution: $ f_(Y|X) (y|x) = (f_(X,Y)(x,y)) / (f_X (x)) $
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- If $f_X (x) > 0, f_(X,Y)(x,y) = f_X (x) f_(Y|X) (y|x)$
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- If $f_X (x) > 0, f_(X,Y)(x,y) = f_X (x) f_(Y|X) (y|x)$
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- $P(Y <= y | X = x) = integral^y_(-infinity) f_(Y|X)(y|x) dif y$
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- $P(Y <= y | X = x) = integral^y_(-infinity) f_(Y|X)(y|x) dif y$
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- $E(Y | X = x) = integral^infinity_(-infinity) y f_(Y|X)(y|x) dif y$
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- $E(Y | X = x) = integral^infinity_(-infinity) y f_(Y|X)(y|x) dif y$
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=== Independent Random Variable
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- *Independent*: $ f_(X, Y)(x, y) = f_X(x) f_Y(y) $
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- *Properties*
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+ If $X, Y$ are independent random variables, $ P(X <= x; Y <= y) = P(X <= x) P(Y <= y) $
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+ $g_1(X) "and" g_2(Y)$ are independent. (E.g. $X^2 "and" log(Y)$ are independent)
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+ if $F_X(x) > 0, "then" f_(Y|X)(y|x) = f_Y (y)$
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#colbreak()
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=== Expectation and Covariance
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- *Expectation*: $ E(g(X, Y)) = sum_x sum_y g(x, y)f_(X, Y)(x, y) \
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E(g(X, Y)) = integral^(infinity)_(-infinity) integral^(infinity)_(-infinity) g(x, y)f_(X, Y)(x, y) dif y dif x $
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#[
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#let cov = "cov"
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- *Covariance*: $ cov(X, Y) = sum_x sum_y (x-mu_x)(y-mu_y)f_(X, Y)(x, y) $
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- *Properties*
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+ $cov(X, Y) = E(X Y) - E(X)E(Y)$
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- $E(X Y) = integral integral x y f(x, y) dif y dif x$
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+ If $X$ and $Y$ are independent, $cov(X, Y) = 0$
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- $X perp Y => cov(X, Y) = 0$
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- $cov(X, Y) = 0 arrow.double.not X perp Y$
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- $E(X Y) = E(X)E(Y)$
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+ $cov(a X + b, c Y + d) = a c dot cov(X, Y)$
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+ $V(a X + b Y) = a^2V(X) + b^2V(Y) + 2 a b dot cov(X, Y)$
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]
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= Probability Distributions
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== Discrete Distributions
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=== Discrete Uniform Distribution
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If random variable $X$ assumes values $x_1, ...$ with _equal_ probability, then $X$ follows discrete uniform distribution. PMF of $X$ is $ f_X(x) = cases(1/k\, &x = x_1\,...\,x_k \ 0 & "otherwise") $
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$ E(X) = sum^k_(i=1)x_i f_X (x_i) = 1/k sum^k_(i=1)x_i $
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$ V(X) = E(X^2) - E(X)^2 = 1/k sum^k_(i=1)x_i^2 - mu^2_x $
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- *Bernoulli Trial*: experiment with only 2 outcomes (1/0)
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- *Bernoulli Random Variable*: X be no of sucess in Bernoulli trial. $X$ has only 2 values. $p$ is probability of success. PMF: $ f_X (x) &= P(X = x) = cases(p &"if" x = 1, 1-p &"if" x = 0) \ &= p^x (1-p)^(1-x), "for" x = 0,1 $
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- $X ~ "Bernoulli"(p)$
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- $q = 1-p$
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- $f_X (1) = p, f_X (0) = q$
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- $E(X) = p$
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- $V(X) = p q$
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- *Bernoulli Process* - repeated independent and identical bernoulli trials
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- *Binomial Random Variable* - No of successes in $n$ trials of bernoulli process.
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- P of $x$ successes in $n$ trials
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- $X ~ "Bin"(n, p)$
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- $P(X = x) = vec(n,x)p^x (1-p)^(n-x)$
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- $E(X) = n p$
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- $V(X) = n p(1-p)$
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- *Negative Binomial Distribution* - No of trials needed for $k$ successes
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- P of $x$ trials needed for $k$ successes
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- $X ~ "NB"(k, p)$
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- $f_(x)(x) = P(X = x) = vec(x - 1, k - 1)p^(k)(1-p)^(x-k)$
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- $E(X) = k/p$
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- $V(X) = ((1-p)k) / p^2$
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- *Geometric Distribution*: No of trials needed until first success occurs.
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- $X ~ "Geom"(p) = "NB"(1, p)$
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- $f_x (x) = P(X = x) = (1-p)^(x-1)p$
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- $E(X) = 1/p$
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- $V(X) = (1-p)/p^2$
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