Stat 330 Formula Sheet Exam II

advertisement
Stat 330
Formula Sheet
Exam II
probability mass function properties: 0 ≤ pX (k) ≤ 1 for all k = 0, 1, 2, . . . and
P
k
pX (k) = 1
probability distribution function properties:
FX (t) non-decreasing for t, limt→−∞ FX (t) = 0, limt→∞ FX (t) = 1, step-function, increases at k.
expected value & variance properties:
E[aX + bY ] = aE[X] + bE[Y ], E[X 2 ] = V ar[X] + (E[X])2 , V ar[aX + b] = a2 V ar[X]
V ar[X + Y ] = V ar[X] + V ar[Y ], iff X, Y are independent.
E[X · Y ] = E[X] · E[Y ] only if X, Y are independent.
P
covariance of X and Y : Cov(X, Y ) = (x,y) (x − E[X])(y − E[y])pX,Y (x, y)
correlation of X and Y : Corr(X, Y ) = √
Cov(X,Y )
,
V ar[X]V ar[Y ]
correlation is in [0, 1], unitless.
discrete distribution(s):
Binomial B(n, p): X = number of successes in n indep. trials, p = P (success) (Bernoulli trials).
pmf: pX (k) = nk pk (1 − p)n−k , for FX (k) see Binomial table, E[X] = np, V ar[X] = np(1 − p)
Geometric Geo(p): X = number of Bernoulli trials until 1st success, p = P (success).
pmf: pX (k) = (1 − p)k−1 p, FX (k) = 1 − (1 − p)k , E[X] = 1/p, V ar[X] =
1−p
p2
Poisson P oi(λ): X = number of events in some unit of time/space,
λ = rate of events in some unit of time/space.
k
pmf: pX (k) = e−λ λk! , for FX (k) see Poisson table, E[X] = λ, V ar[X] = λ
continuous random variable X uncountable set of possible outcomes (=interval),
probability density function : fX (k) = F 0 (X = k),
cumulative distribution function FX (t) = P (X ≤ t),
R∞
R∞
expected value E[X] = −∞ xfX (x)dx, variance V ar[X] = −∞ (x − E[X])2 fX (x)dx.
R∞
probability density function (pdf ) properties: fX (x) ≥ 0 for all x and −∞ fX (x)dx = 1.
cumulative distribution function (cdf ) properties:
FX (t) non decreasing for t, limt→−∞ FX (t) = 0, limt→∞ FX (t) = 1.
rules for densities & distribution functions
Rt
Rb
P (a ≤ X ≤ b) = a fX (x)dx = FX (b) − FX (a), P (X = a) = 0, FX (t) = −∞ fX (x)dx.
Uniform U (a, b), X = random value between a and b, all ranges of values in [a, b] are equally likely,
pdf: f (x) = 1/(b − a) for x ∈ [a, b], cdf: FX (x) =
x−a
b−a
for x ∈ [a, b],
E[X] = (a + b)/2, V ar[X] = (b − a)2 /12.
Exponential Exp(λ), X = time/space until 1st occurrence of event,
λ = rate of events in some unit of time/space.
pdf: f (x) = λe−λx for x ≥ 0, FX (x) = 1 − e−λx for x ≥ 0,
E[X] = 1/λ, V ar[X] = 1/λ2 .
Exponential distribution is memoryless, i.e. P (X ≤ s + t | X ≥ s) = P (X ≤ t).
Erlang Distribution Erlang(k, λ): If Y1 , . . . , Yk are k independent exponential random variables
with parameter λ, their sum X has an Erlang distribution:
P
X := ki=1 Yi is Erlang(k, λ) k is stage parameter, λ is rate parameter
Erlang density f (x) = λe−λx ·
(λx)k−1
(k−1)!
for x ≥ 0
E[X] = k/λ, V ar[X] = k/λ2
Erlang cdf is calculated using Poisson cdf: FX (t) = 1 − FY (k − 1)
where X ∼ Erlang(k, λ) and Y ∼ P oi(λt)
so use Poisson cdf table with parameter= λt
Normal r.v.: X ∼ N (µ, σ 2 ), Normal density is “bell-shaped” f (x) =
√ 1 e−
2πσ 2
(x−µ)2
2σ 2
E[X] = µ, V ar[X] = σ 2 .
standardization: FX (x) = Φ( x−µ
σ ); Z ∼ N (0, 1) and Φ(z) ≡ FZ (z) and Φ(−z) = 1 − Φ(z).
2 )
X ∼ N µx , σx2 ), Y ∼ N (µy , σy2 ), then W := aX + bY has normal distribution W ∼ N (µW , σW
2 = a2 σ 2 + b2 σ 2 + 2abCov(X, Y )
where µW = aµx + bµy and σW
x
y
Central Limit Theorem (CLT): If X1 , X2 , . . . , Xn are i.i.d. r.v.’s with E[Xi ] = µ, V ar[Xi ] = σ 2 ,
P
P
approx
approx
then X := n1 ni=1 Xi ∼ N (µ, σ 2 /n) and Sn = i Xi ∼ N (nµ, nσ 2 )
Bin(n, p)
P oi(λ)
approx
∼
approx
∼
N (np, np(1 − p)) for large n (if np > 5),
N (λ, λ) for large λ,
Download
Random flashcards

Arab people

– Cards

Pastoralists

– Cards

Radioactivity

– Cards

Nomads

– Cards

Create flashcards