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Important Formulas for Statistics Basic stuff πππ(π) = πΆππ£(π, π) = πΈ[(π β π)(π β π)] = πΈ[(π β π)2 ] = πΈ[π 2 ] β (πΈ[π])2 Variance Properties Expected value Properties Is a linear operator, E[XY]=E[X]E[Y] if independant. Limit in probability A sequence {Xn} of random variables converges in probability towards X if for all Ξ΅ > 0 Properties of prob. convergence Gaussian Integral Convergence in probability implies convergence in distribution , x can be a linear function of the integrand variable Joint, Marginal and Conditional probability Expectation and var conditional Conditional Probaibility = π(π,π) π(π) Sum of a RVits mean β(π₯π β π₯Μ ) = 0 Sample Variance Prop of log Liebniz rule on the differential of an integral Binomial Coefficient (helpful in convolution of poisson) Expectation (useful for the proof of Rao Blackwell thm) in conditional Transformations One Variable (discreet) One Variable (continuous) Two Variables Jacobian Order Statistics π[πΉ(π₯)]πβ1 π(π₯) Maximum In general Same distribu. Min Between two (diff distribution) = Between min and max Idea behind multiplication of supports π(π β 1)[πΉ(π¦) β πΉ(π₯)]πβ2 π(π₯)π(π¦) If 1(xi<ΞΈ) then multiplication can be summarized in 1(x(n)<ΞΈ). Conversely if 1(xi<1/ΞΈ)ο 1(ΞΈ <1/x(n)). Convolutions Always check that the supports are compatible to establish the limits of the integration/summation Remember that in a convolution the support can be seen as a straight line crossing the plane in (z,0) and (0,z). (e.g. when adding two uniforms the limits of integration vary when z>1) Sufficiency Sufficiency (I) Conditional distribution approach (By definition) ππ (π = π₯|π = π‘) = π(π = π₯|π = π‘). That is this pmf/pdf is free of ΞΈ. π(π = π₯) β π(π = π‘|π = π₯) π(π = π₯) ππ (π = π₯|π = π‘) = = β 1(π(π₯) = π‘) π(π = π‘) π(π = π‘) β² ππ|π=π‘ (π₯) ππππ π π‘πππ£πππ£π π Sufficiency (II) Neyman πΉπππ‘ππππ§ππ‘πππ: πΏ(π, π₯) = β(π₯)π(π(π₯), π)ο T(x) is suff. Factorization theorem Hint for proof: Dicreete case onlyο : X is contained in T(x). Then make g(T(x),ΞΈ)=PΞΈ(T(X)=T(x)). ο: Use the distribution of T:Pπ (T(X) = t). = β β:π(β)=π‘ π¦ πΏ(π) to divide in π¦ πΏ(π, π₯) , to ultimately find that PΞΈ(X=x|T(X)=t) doesnβt depend on ΞΈ. Minimal sufficiency: Lehmann Scheffé (also β(π₯, π₯π‘ ; π) = πΏ(π,π₯) πΏ(π,π₯π‘ ) β π(π₯) = π(π₯π‘ ) π . π‘. β ππ ππππ ππππ π β π ππ ππππ π’ππ πππ π. works with non identically If we T(x)=T(y) only happens for x=y then we can look at the order distr) statistics. Prof: The sufficiency is with a partition and the β, the minimal is with β and using the fact that T is a function of another sufficient. Information Helpful in: CRLB, Asymptotic MLE Arcillarity Good trick: If the distribution of Y=ΞΈX does not involve ΞΈ, then T=Xi/Xj, doesnβt either. Families i)Location, ii)Scale iii)Locationscale. Ancillary in each family is easy to achieve by subtraction and/or division by another element of the sample (or an order stat) Completeness A stat both Complete and A function 1-1 of a minsuff is a minsuff sufficient ο minsuff Basus theorem: w is complete and sufficient for ΞΈ, U is ancillary, ο U and W are independent Hint proof: WTS: Conditional distribution of U on W is the distribution of U. Then use completeness using these distributions as functions. UMVUE and Estimators UMVUE: Smallest variance in the family of unbiased estimator. How to find estimators? Method of moments Equate the theoretical moments to the sample ones Maximum likelihood 1. Find the L 2. Ln(L) 3.Maximize (and check is maximum) using FOC and SOC Invariance of MLE You can apply a function to the MLE st you get the MLE of the function How to compare them? Mean Squared error Bias Μ) β π½ B(π½)=π¬(π½ BLUE Sumatory (same weight) of the unbiased estimators How to improve them? Rao Blackwell (improve) T is an unbiased est. of Ο(ΞΈ), U is (jointly) sufficient for ΞΈ. π = π(π’) = πΈπ’ [π|π = π’]. W is unbiased for Ο, and has lower variance than T. Proof: Use Confitional expectation to check that W is unbiased, 2 and use the Var(T)=E[(T- Ο(ΞΈ)) ] with W on the middle. CRLB (compare) Hint proof: Use chain rule of Οβ(ΞΈ) so Οβ(ΞΈ)=Cov(TY)<V(T)V(Y) then equate with the V(Y)=nE(β¦) Lehman Scheffé (find) T is an unbiased est. of Ο(ΞΈ), U is (jointly) complete and sufficient for ΞΈ. π = π(π’) = πΈπ’ [π|π = π’]. W is UMVUE. Hint proof: Suppose W* another UMVUE , use completeness to prove W*=W βHow to apply? 1) Find a complete and suff stat for ΞΈ 2) Find a function of this compsuff stat that is unbiased for the thing you are trying to estimate. Sometimes the function could be a simple multiplication or sum of 1) THE EXPONENTIAL FAMILY (and reasons to love it) π(π₯, π) = π(π) β π(π₯) β π π(π)βπ (π₯) 1. We can Easily find the Expected value πΈπ (π (π₯)) = βπβ²(π) , π(π)πβ²(π) 2. T(x)=Ξ£R(xi) is minsuff and complete. 3. It follows the MLR property Tests Errors Simple vs simple: Neymann Pearson Lemma 1)Define the test STEPS 2)Find the Likelihood ratio 3) Choose the statistics of interest and define Rejection region: An MP a level test always depend on the sufficient statistic. 4)Operate to a form that is implementable either by knowing the distribution of the statistic (take it to chi or normal) or by using π ππ0 (π ) = πΌ (e.g. β«π 0 π(π₯)ππ₯ Remember that if the statistic of interest is decreasing because of the other constants then k must behave properly. (eg. In a normal when µ0< µ1, then the rejection is (β¦)>z, but if µ0> µ1 then the rejection is (β¦)<-z. One side composite UMP via Neymann Person Fix a value ΞΈ1>ΞΈ0. Then we can find the simple vs simple, and say that if it doesnβt matter which value of theta we choose it also applies here. MLR property One side composite with or MLR property (Karlin Rubin) If MLR Non decreasing: IF MLR non increasing: With T(x) being a suff stat. GLRT 1)Define the Rejection region 2)Calculate 3) The sup(β¦) is the MLE, so in the case in which the null hypothesis is simple then is simply replacing, while for the alternative we shall use the MLE. 4)Operate to a form that is implementable by operating the K with the KNOWN variables, that is express the test in terms of the MLE and the parameter of the null. Check if the exponents are making a two sided inequality. π 5)Apply ππ0 (π ) = πΌ (e.g. β«π 0 π(π₯)ππ₯ ) Asymptotic FOR MLE Delta Method You can then operate the left inside affecting the right as regular normal. E.g. multipliying by a constant will multiply the µ but will be squared with the Ο You can also apply the delta method in this case, in which case you arrive to MLE being Efficient (by CRLB). Confidence intervals Assymptotic: