Formulas

Discrete:

Continuous:

E(X)=p

Var(X)=pq

E(X)=np

Var(X)=npq

E(X)=1/p

Var(X)=q/(p)^2

E(X)=μ

Var(X)=σ^2

E(X)=n/λ

Var(X)=n/λ^2

where

S^2

N(µ, σ^2/n)

either factorization theorem or

show that conditional prob. of X given Y=y is not dependent on theta

if Y is sufficient for theta and for sufficient statistic Z there exists function r so that Y=r(Z)

or

find other sufficient statistic Z and if you cant write Y as Z, THEN Y is not MSS

complete & sufficient-> MSS

nonconstant & complete -> not ancillary

constant -> ancillary & complete

… then Y is MSS for θ

(Bahadur)

… then for all θ ∈ Θ, Y and Z are independent with respect to fθ.

(Basu)

sum over x E(Y|X=x)P(X=x)

Let Z be a sufficient statistic for {fθ : θ ∈ Θ} and let Y be an estimator for g(θ). Define W := Eθ(Y |Z)

ell(θ, y) is convex

Y is uniformly minimum risk unbiased (UMRU) for risk function r if, for any other unbiased estimator Z for g(θ), we have

Y is uniformly minimum variance unbiased (UMVU) if, for any other unbiased estimator Z for g(θ), we have

Let {fθ : θ ∈ Θ} be a family of distributions and let W be an unbiased estimator for g(θ). Let L2(Ω) be the set of statistics with finite second moment. Then W ∈ L2(Ω) is UMVU for g(θ) if and only if Eθ(WU) = 0 ∀ θ ∈ Θ, for all U ∈ L2(Ω) that are unbiased estimators of 0.

also: Var(Y)=

Let Z be a complete sufficient statistic for {fθ : θ ∈ Θ} and let Y be an unbiased estimator for g(θ). Define W := Eθ(Y |Z). Assume that l(θ, y) is convex in y, for all θ ∈ Θ. Then W is UMRU for g(θ). If l(θ, y) is strictly convex in y for all θ ∈ Θ, then W is unique.

In particular, W is the unique UMVU for g(θ).

1. find joint pdf

2. find marginal pdf

3. find conditional pdf

4. find conditional expectation

Var(X)+(E(X))^2

Y := t(X_1,...,X_n)

Y is a sufficient statistic for θ if, for every y ∈ Rk and for every θ ∈ Θ, the conditional distribution of (X_1, . . . , X_n) given Y = y (with respect to probabilities given by fθ) does not depend on θ

Calculate whatever moment that we want to find estimator for (first moment = EX, second moment = EX^2)

rewrite result so that we have: θ=c EX where c is a constant

lets say c= 3/2

the method of moments estimator is then:

no but they are conditionally independent when conditioned on X1,…X_n

function:

derivative:

is equal to the variance of the estimator

Let X1, . . . , Xn ∼ U(0, θ) be an iid sample, where θ is unknown