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G(n)=bx (lnb)n So bx = X1 n=0 g(n)(0) n!(e) the variance of Y 4 Let Y be a random variable having mean µ and suppose that E(Y −µ)4 ≤ 2 Use this information to determine a good upper bound to P(Y −µ ≥ 10) 5 Let U and V be independent random variables, each uniformly distributed on 0,1 Set X = U V and Y = U − V Determine whether or not X and Y areG & ¯ L }7 è à Æ ¹ ñ µ ñ ® £ Ö ¢ s?

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1 (n¡‚)x {z }!1 µ 1¡ ‚ n ¶n {z }!e¡‚ = e¡‚‚x x!Q d V b u O ð E ñ X ì E ñ X ì ¸ S y ³ ß Ú ¡ ³ y è ­ W ö O 6'0$ y ² C 6'0$ W Í & v Ï ¤ ¤ J G/ ¥ b q O Ö û = õ W > b q O F õ S W M D ¦ ­ ê ¿ É ñ ó U W Z ½ RU å ½ PJ G/ r M n q &' VWDJH s u d i s u } * Ú y Ñ &' VWDJH r Í & vµ θσ 2 /2 1 = EX = Ee Y = M Y (1) = e µ 2θ2σ 2 2 = EX 2 = Ee 2Y = M Y (2) = e (b) First, note that µ 2 σ 2 2 /(µ 1) = e It follows that a methodofmoments estimate for σ 2 is σˆ 2 = ln(ˆµ 2 /µˆ 2 1) where µˆ 1 = 1 n X i n i =1 µˆ 2 =

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E(Y) = E E(YjX) = Z E(YjX= x)p X(x)dx The Law of Total Variance is Var(Y) = Var E(YjX) E Var(YjX) The moment generating function (mgf) is M X(t) = E etX If M X(t) = M Y(t) for all tin an interval around 0 then X =d Y The moment generating function can be used to \generate" all the moments of a distribution,Y n m Í o%3, Þ è s b q O d } j Ü ê µ r z I T y n m Í o ¦ " / g s b q O d } ¢ ñ Ç r z È ¢ ñ Ç < y T L r M Þ é ª Á Ï ® é Õ s y û Æ Â ê c q y n m È V b j è Ä é ¤ Ö y U q O d }1 ra ndom v ector with mean µ x and v aria nce co v ar iance ma trix !

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(b) g(x)=cosx Solution (a) We observe that f =sinx =) f (0)=0;Math 541 Statistical Theory II Methods of Evaluating Estimators Instructor Songfeng Zheng Let X1;X2;¢¢¢; be n iid random variables, ie, a random sample from f(xjµ), where µ is unknown An estimator of µ is a function of (only) the n random variables, ie, a statistic ^µ= r(X 1;¢¢¢;)There are several method to obtain an estimator for µ, such as the MLE,N È Ï n ½ = ¹ Â w ½ z z µ Â ¡ þ C C Æ b A C Æ e ° V ÿ ` i \ þ Á 1 7 1 B L Î e Á ~ \ Ì ±¹¹° ³ \ þ þ ) â e µ ÿ Á ) â ÿ z ³ \ þ Á B Ò Ì þ²°°´ ³ \ Ì ÿ 8 c þ Á 1 » 1 K Á \ ð M Á Q é Á ò

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X ismultivariatenormal⇔ a′x isnormalforalla def'n x ∼ Np(µ,Σ) ⇔ a′x ∼ N(a′µ,a′p thm If x ∼ Np(µ,Σ) then its characteristic function is φx(t) = exp(it′µ− 1 2t ′Σt) Proof Let y = t′x Then the cf of y is φy(s) def= E{eisy} = exp{isE(y)−1 2s 2var(y)} = exp{ist′µ−1 2s 2t′Σt} Then the cf of x~ s k ¯ oB ñ S E ¥ T Å ' Í ¥% E p ¥( ' ^ B Õ V õ v" ñ $ ?

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Í Ç Ò Æ Å Î É?2 Note that approximation works better when n is large and p is small as can been seen in the following plot If p is relatively large, a difierent approximation should be used This is coming later (Note(x−µ)2 2 ˙ is the N(µ,1) density, and f 2(xµ,τ) = 1 √ 2πτ2 exp ˆ − (x−µ)2 2τ2 ˙ is the N(µ,τ2) density Then the expectation of a random variable with this mixture density is given by EX i = Z ∞ −∞ xf(xµ,τ2,p)dx = Z ∞ −∞ x pf 1(xµ)(1−p)f 2(xµ,τ2) dx

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½Þ q z » å!BASIC STATISTICS 5 VarX= σ2 X = EX 2 − (EX)2 = EX2 − µ2 X (22) ⇒ EX2 = σ2 X − µ 2 X 24 Unbiased Statistics We say that a statistic T(X)is an unbiased statistic for the parameter θ of theunderlying probabilitydistributionifET(X)=θGiventhisdefinition,X¯ isanunbiasedstatistic for µ,and S2 is an unbiased statisticfor σ2 in a random sample 3Stat 110 Final Review, Fall 11 Prof Joe Blitzstein 1 General Information ThefinalwillbeonThursday12/15, from2PMto5PMNobooks, notes, computers,

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^ q b ( n j y & Ú } t z Ö W à c k n j y V ö Ñ R d y v J Þ è V U Ü j v õ ¢ r / b j Q k } n q ¦ y 0 Ô À n à \ b % v ` n j ÿ b O Ñ / k n j } & Ú z µ \ \ t F ( Ï r & Ú µ O j y ° S q O } { O ° o ¨ ñ / r Î y Ø Á ­ r(b) (7 points) Derive , the variance of U, in terms of b, and the covariance 2 σU 2, 2 σX σY σXY 2 σU = EU 2 − E(U)2 = EU2 because the second term is zero = E(Y − µ Y) 2 − 2b(X−µ x)(Y − µ Y) b 2(X−µ x) 2 = E(Y − µ Y) 2 − 2bE(X−µ x)(Y − µ Y) b 2E(X−µ x) 2 = 2 − 2b σY σXY b 2 2 σX (c) (6 points) Suppose I want to choose b in order toµ X = EX = Z ∞ −∞ xf X(x) dx The expected value of an arbitrary function of X, g(X), with respect to the PDF f X(x) is µ g(X) = Eg(X) = Z ∞ −∞ g(x)f X(x) dx The variance of a continuous rv Xwith PDF f X(x) and mean µ X gives a quantitative measure of how much spread or dispersion there is in the distribution of xvalues The

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µ µ µ µ µ µ µ Å µ z ¶ N µ z ¶ z y à ß µ µ q µ / u q N 2 { q z ~ » B Þ # 6 Þ å!Ten as E(X) or µ X If we observe N random values of X, then the mean of the N values will be approximately equal to E(X) for large N The expectation is defined differently for continuous and discrete random variables Definition Let X be a continuous random variable with pdf f X(x) The expected value of X is E(X) = Z ∞ −∞ xf XExample 54 Find Maclaurin series for (a) f (x)=sinx;

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Y (t), then Z = X Y has the moment generating function, M Z(t) = M X(t)M Y (t) 2 Find a variance of the random variables in Example 1 Finally, we can also define the conditional expectation, E(X Y), and conditional variance, E(X− µ X)2 Y), of a random variable X given another random variable Y The expectation is over theõ Ö ÿ n ^ d b { ¿ ï 3 s r t b ;µλ,x 1n ∼N(M λ,L −1 λ) where L λ= λ 0 nλ and M λ= λ 0µ 0 λ P n i=1 x i λ 0 nλ Foranyfixedvalueofµ,itisstraightforwardtoderive3 that λµ,x 1n ∼Gamma(A µ,B µ) (1) whereA µ= a n/2and B µ= b 1 2 P (x i −µ)2 = nˆσ2 n(¯x −µ)2 where ˆσ2 = 1 n P (x i

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