0000030318 00000 n The goal of this lecture is to explain why, rather than being a curiosity of this Poisson example, consistency and asymptotic normality of the MLE hold quite generally for many \\ 0000005954 00000 n converges in distribution to a normal distribution (or a multivariate normal distribution, if has more than 1 parameter). (15) \hat{\theta}_N \rightarrow^d \mathcal{N}(\theta_0, \mathcal{I}_N(\theta_0)^{-1}). (23) First, we will show, ee=M,(13) L^{\prime\prime}_N(\theta) &= \frac{\partial^2}{\partial \theta^2} \left( \frac{1}{N} \log f_X(x; \theta) \right). Ill then show how these assumptions imply some established properties of the OLS estimator ^\hat{\boldsymbol{\beta}}^. /BBox[0 0 16 16] \hat{\theta}_N \rightarrow^d \mathcal{N}(\theta_0, \mathcal{I}_N(\theta_0)^{-1}). \\ (28) 0000002391 00000 n \\ \mathbf{M} = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}, \tag{14} 0 &= \sum_{n=1}^N \left[ \frac{X_n}{p} + \frac{X_n}{1 - p} \right] - \frac{N}{1 - p} \end{cases} \tag{22} Assumption 111 is just Equation 111; it means that we have correctly specified our model. \tag{9} to the proof of unbiasedness. Obviously, one should consult a standard textbook for a more rigorous treatment. (20) \tag{6} \\ Lets tackle the numerator and denominator separately. &= \mathbf{X} - \mathbf{H} \mathbf{X} (10) /Type/XObject Then, we apply our variance reduction method by choosing optimally the combination weight in the redened dependent variable. stream /Type/XObject 0000023921 00000 n (23) 0000003375 00000 n cEUNF`.t&XuAuB"A(F3Nis6 XUjiZKxy|k:])h*f$}P|z"_u"xTy|8lW$@$w%uTWYubjyhZ[Ih 4yl5WkL-2E{)L\gBS+Moka.4z{ (-Z~jYST2T6U64Er)HVQeq55HMfr~UA)4F&~~:oWgj%zU.W>?wOJq{:6W_NSua_.K[7/|@:Zn QyT9Mo||VAZ`!w|uDmL .sju{.iUL~mMtWnu\&_?==UVw`X6XWpK2E4_][ ~=~#L-/ue^R:ZEqjG. Comparison of consistency versus unbiasedness. (22), The Fisher information is the negative expected value of this second derivative or, IN(p)=E[n=1N[Xnp2+Xn1(1p)2]]=n=1N[E[Xn]p2E[Xn]1(1p)2]=n=1N[1p+11p]=Np(1p). (2) 0000004823 00000 n without assuming A.MLR6 and so without assuming normal errors: #1 If i = 0; then b i =se b i is asymptotically N(0;1): Recalling that a Student-t approaches a normal as the number of degree of freedom approaches in-nity, it is true that for n large . (9) If OLS estimators satisfy asymptotic normality, it implies that: a. they have a constant mean equal to zero and variance equal to sigma squared. 0 %%EOF N(0;Q2P): The value for could also be obtained along a dierent route. samples from a Bernoulli distribution with true parameter ppp. \\ % &= \mathbb{E}\left[ 9 0 obj 0000010869 00000 n \\\\ First, lets compute the derivative: plogfX(X;p)=n=1N[pXnlogp+p(1Xn)log(1p)]=n=1N[Xnp1Xn1p]=n=1N[Xnp+Xn11p]. 0) 0 E( = Definition of unbiasedness: The coefficient estimator is unbiased if and only if ; i.e., its mean or expectation is equal to the true coefficient Taken together, we have, LN(~)pE[22logfX(X1;0)]=I(0). 0000024472 00000 n Variances and standard errors in large samples. &= N - \text{trace}\left( \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \right) Here is the minimum code required to generate the above figure: I relied on a few different excellent resources to write this post: # Plot the asymptotically normal distribution. 0000008158 00000 n 0000004483 00000 n Proposition If Assumptions 1, 2, 3 and 4 are satisfied, then the OLS estimator is asymptotically multivariate normal with mean equal to and asymptotic covariance matrix equal to that is, where has been defined above. Then under the conditions of Theorem 27.1, if we set n = n '0 . &= \sigma^2 \mathbf{A A}^{\top} XN(0,2IN),(25). \sqrt{N}(\hat{\theta}_N - \theta_0) = - \frac{\sqrt{N} L_N^{\prime}(\theta_0)}{L_N^{\prime\prime}(\tilde{\theta})} \tag{7} c. they are approximately normally distributed in samples with less than 10 observations. For the numerator, by the linearity of differentiation and the log of products we have, NLN(0)=N(1N[logfX(X;0)])=N(1N[logn=1NfX(Xn;0)])=N(1Nn=1N[logfX(Xn;0)])=N(1Nn=1N[logfX(Xn;0)]E[logfX(X1;0)]). \mathbb{E}[s^2 \mid \mathbf{X}] &= \sigma^2, 51 0 obj <> endobj This section is not strictly necessary for understanding the sampling distribution of ^\hat{\boldsymbol{\beta}}^, but its a useful property of the finite sample distribution, e.g. OLS makes a few important assumptions (assumptions 111-444), which mathematically imply some basic properties of the OLS estimator ^\hat{\boldsymbol{\beta}}^. In this model, the asymptotic variance of the OLS estimator has even remained equal to the variance of the correctly specied model. \end{aligned} Therefore, IN()=NI()\mathcal{I}_N(\theta) = N \mathcal{I}(\theta)IN()=NI() provided the data are i.i.d. V[nX]=2,n{1,,N}.(3). Tm \tag{6} Proof of asymptotic normality To prove asymptotic normality of MLEs, define the normalized log-likelihood function and its first and second derivatives with respect to as 0000006973 00000 n This is not bad. V[X]=2IN.(5). &= \frac{1}{N} \left( \frac{\partial^2}{\partial \theta^2} \log f_X(X; \theta) \right) The goal of this post is to discuss the asymptotic normality of maximum likelihood estimators. %%EOF &= N - P. \\ Introductionhttps://youtu.be/xZ_-xRWSVZsAsymptotic Theory for . 0000011843 00000 n We can then use those assumptions to derive some basic properties of ^\hat{\boldsymbol{\beta}}^. \mathbb{E}\left[\frac{\partial}{\partial \theta} \log f_X(X_1; \theta_0)\right] = 0. (7) 0000028280 00000 n (18) Our theorem on asymptotic normality also implies that, regardless of the error's distribution, the OLS estimators, when properly standardized, have approximate standard normal distributions. \tag{2} As the asymptotic results are valid under more general conditions, the OLS MX=(INH)X=XHX=XX(XX)1XX=0.(19). &\stackrel{\dagger}{=} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} (\mathbf{X} \boldsymbol{\beta} + \boldsymbol{\varepsilon}) - \boldsymbol{\beta} endstream endobj startxref This proof uses basic properties of the trace operator: trace(M)=trace(INH)=trace(IN)trace(H)=Ntrace(X(XX)1X)=Ntrace(XX(XX)1)=Ntrace(IP)=NP. 2. |nxMPf kw6to51=5jg!HRddZHwIP&"$HLV16=-=qFT$)2si~2vf$cdzbn$|qHddi?#`>?tn"S& k#BNii532He IZg5###uk$Ce@ N? \begin{aligned} \\ \tag{7} &= \mathcal{I}(\theta_0). \end{aligned} \tag{24} E[nX]=0,n{1,,N}.(2). This proof is from (Hayashi, 2000). /Filter/FlateDecode 0000028323 00000 n \boldsymbol{\varepsilon} \sim \mathcal{N}(\mathbf{0}, \sigma^2 \mathbf{I}_N). \\ &= \mathbb{E}\left[ (21), In other words, the MLE of the Bernoulli bias is just the average of the observations, which makes sense. \boldsymbol{\varepsilon} \mid \mathbf{X} \sim \mathcal{N}(\mathbf{0}, \sigma^2 \mathbf{I}_N), \tag{25} Abbott PROPERTY 2: Unbiasedness of 1 and . \mathbf{y}^{\top} \mathbf{M} \mathbf{X} \boldsymbol{\beta} <]>> \mathbb{E}[\hat{\boldsymbol{\beta}} - \boldsymbol{\beta} \mid \mathbf{X}] = \mathbf{0}. LN(^N)=LN(0)+LN(~)(^N0). \sqrt{N} L^{\prime}_N(\theta_0) Finally, note that Equation 888 means we can write write the sampling error in terms of the residuals: ^=(XX)1X. /FormType 1 Then there exists a point c(a,b)c \in (a, b)c(a,b) such that, f(c)=f(a)f(b)ab(5) f^{\prime}(c) = \frac{f(a) - f(b)}{a - b} \tag{5} f(c)=abf(a)f(b)(5), where f=LNf = L_N^{\prime}f=LN, a=^Na = \hat{\theta}_Na=^N and b=0b = \theta_0b=0. stream (24) &\stackrel{\ddagger}{=} \mathbf{A} \mathbb{V}[\boldsymbol{\varepsilon} \mid \mathbf{X}] \mathbf{A}^{\top} (12) endstream /Resources 10 0 R 1) 1 E( =The OLS coefficient estimator 0 is unbiased, meaning that . /Type/XObject \\ By other regularity conditions, I simply mean that I do not want to make a detailed accounting of every assumption for this post. pK{wLW 0000003541 00000 n ]*hYBApc =-AWOJ}k~6W_m@rPge^AT*u'.m) C1-%7|yY1_N"">Lo6vZ} 'j;kJi"BXw]Zb_0y;a{`/=RyKBCZ>JEx$nHl7zwN7KO`NdQ1pZ9\\*$vLY9*5-zOg]w3G'N\ssXSiARJI^Fx)a6Cz1_YBh6$EQ3!CJBHjqrhbcFh6x!t4-OVVh :y>i3sVmxZ86;.ecPWB*j7{XXdByf$mzGpCrRi#. xS0PpW0PHWP( This property of OLS says that as the sample size increases, the biasedness of OLS estimators disappears. /FormType 1 Consistency. &= \mathbb{E}\left[\left(\frac{\partial}{\partial \theta} \log f_X(X_1; \theta_0)\right)^2\right] - \left(\underbrace{\mathbb{E}\left[\frac{\partial}{\partial \theta} \log f_X(X_1; \theta_0)\right]}_{=\,0}\right)^2 If youre unconvinced that the expected value of the derivative of the score is equal to the negative of the Fisher information, once again see my previous post on properties of the Fisher information for a proof. \\ \mathbb{E}[\boldsymbol{\varepsilon}^{\top} \mathbf{M} \boldsymbol{\varepsilon} \mid \mathbf{X}] Our claim of asymptotic normality is the following: Asymptotic normality: Assume ^Np0\hat{\theta}_N \rightarrow^p \theta_0^Np0 with 0\theta_0 \in \Theta0 and that other regularity conditions hold. In statistics, a regression model is linear when all terms in the model are either the constant or a parameter multiplied by an independent variable. &\rightarrow^p \mathbb{E}\left[ \frac{\partial^2}{\partial \theta^2} \log f_X(X_1; \theta) \right]. 1 Answer. Econometrics 3 5.1 Consistency Under the Gauss-Markov assumptions OLSE . If we compute the derivative of this log likelihood, set it equal to zero, and solve for ppp, well have p^N\hat{p}_Np^N, the MLE. This proof is also from (Hayashi, 2000), but Ive organized and expanded it to be more explicit. E[^X]=0.(7). Asymptotic Theory for Consistency Consider the limit behavior of asequence of random variables bNas N.This is a stochastic extension of a sequence of real numbers, such as aN=2+(3/N). Asymptotic Normality and Rate of Convergence in Distribution Sense We shall give a theorem the proof which is based on Sacks' Theorem 17 [ 102] (see Appendix A ). (3) b. they are approximately normally distributed in large enough sample sizes. Assumptions 111 and 333 are not terribly interesting here. s^2 = \frac{\mathbf{e}^{\top} \mathbf{e}}{N - P}, \tag{11} To do this, we need to make some assumptions. \mathbf{M}\mathbf{X} &= (\mathbf{I}_N - \mathbf{H}) \mathbf{X} If we assume normality in the errors, then clearly, XN(0,2IN),(25) >> (10) Assumption 222, strict exogeneity, is that the expectation of the error term is zero: E[nX]=0,n{1,,N}. \begin{aligned} Then for some point c=~(^N,0)c = \tilde{\theta} \in (\hat{\theta}_N, \theta_0)c=~(^N,0), we have, LN(^N)=LN(0)+LN(~)(^N0). A linear function of a normal random variable \boldsymbol{\varepsilon} is still normally distributed, meaning that ^\hat{\boldsymbol{\beta}} - \boldsymbol{\beta}^ is normally distributed. NLN(0)dN(0,I(0))(14) 0000001754 00000 n Jitty Asks: Asymptotic normality of OLS estimators in practice Apologies in advance because I notice quite a few questions that have similar titles but I didn't see one that answered my specific curiousity: Supposedly OLS coefficient estimators are asymptotically normal, regardless of the. 0000024151 00000 n \middle| \mathbf{X} \right] trailer E[s2X]E[eeX]=2,=(NP)2.(12), We will prove this three step. ^N0=LN(~)LN(0)N(^N0)=LN(~)NLN(0)(7). 0000003506 00000 n 2.1. mean zero, not normal, but with common finite variance. \\ \\ Proof of asymptotic normality of Maximum Likelihood Estimator (MLE) Ask Question Asked 4 years, 9 months ago. 0000008416 00000 n V[^X]=V[^X]=V[(XX)1XAX]=AV[X]A=A(2IN)A=2AA=2(XX)1.(10). 0000048978 00000 n &= \sum_{n=1}^N \left[ X_n \log p + (1 - X_n) \log (1 - p) \right]. \hat{\boldsymbol{\beta}} - \boldsymbol{\beta} \sim \mathcal{N}(\mathbf{0}, \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1}). endstream \text{trace}(\mathbf{M}) &= \text{trace}(\mathbf{I}_N - \mathbf{H}) &\stackrel{\dagger}{=} \mathbf{y}^{\top} \mathbf{M} \mathbf{M} \mathbf{y} Assumption 333 is that our design matrix X\mathbf{X}X is full rank; this property not relevant for this post, but I have another post on the topic for the curious. The standard assumptions of OLS are: Linearity Strict exogeneity No multicollinearity Spherical errors Normality (optional) Assumptions 1 and 3 are not terribly interesting here. fb @7QDE~(1B0hBJ! \end{aligned} \tag{20} (21) \\ Therefore, a low-variance estimator ^N\hat{\theta}_N^N estimates the true parameter 0\theta_00 more precisely. /Length 23 /Subtype/Form 01pN1pp(1p)=n=1N[pXn+1pXn]1pN=n=1NXn[p1+1p1]=N1n=1NXn.(20). &= \frac{1}{N} \left( \frac{\partial^2}{\partial \theta^2} \log \prod_{n=1}^N f_X(X_n; \theta) \right) Property 5: Consistency One of the main uses of the idea of an asymptotic distribution is in providing approximations to the cumulative distribution functions of statistical . /Filter/FlateDecode To state our claim more formally, let X=X1,,XNX = \langle X_1, \dots, X_N \rangleX=X1,,XN be a finite sample where XP0X \sim \mathbb{P}_{\theta_0}XP0 with 0\theta_0 \in \Theta0 being the true but unknown parameter. \\ This post relies on understanding the Fisher information and the CramrRao lower bound. 910 0 obj <> endobj {4F%%X@yVB%u0_,4q 4:rl9#MKZ:::L. Now lets set it equal to zero and solve for ppp: 0=n=1N[Xnp+Xn1p]N1pN1p=n=1NXn[1p+11p]p(1p)1p=1Nn=1NXn. Asymptotic normality Sketch of proof: I follows by arguments similar to our derivation of the delta method. &= \sum_{n=1}^N \left[ \frac{X_n}{p} - \frac{1 - X_n}{1 - p} \right] The asymptotic process is integrated with either one unit root (located at 1 or $-1$), or even two unit roots located at 1 and $-1$. E[M]=j=1Ni=1NMjiE[jiX]. 0000029880 00000 n \begin{aligned} As our finite sample size NNN increases, the MLE becomes more concentrated or its variance becomes smaller and smaller. The log likelihood is, logfX(X;p)=n=1Nlog[pXn(1p)1Xn]=n=1N[Xnlogp+(1Xn)log(1p)]. 51 54 ^NdN(0,IN(0)1).(2). \\ usual t-statistic is asymptotically normal &\stackrel{*}{=} \mathbf{A} (\sigma^2 \mathbf{I}_N) \mathbf{A}^{\top} (19) \tag{27} \sum_{j=1}^N \sum_{i=1}^N M_{ji} \varepsilon_j \varepsilon_i NLN(0)dN(0,V[logfX(X1;0)]).(10). Hence, asymptotic properties of OLS model are discussed, which studies how OLS estimators behave as sample size increases. Another property that we are interested in is whether an estimator is consistent. By definition, the MLE is a maximum of the log likelihood function and therefore, ^N=argmaxlogfX(x;)LN(^N)=0. \\ My in-class lecture notes for Matias Cattaneos. L_N^{\prime}(\hat{\theta}_N) = L_N^{\prime}(\theta_0) + L_N^{\prime\prime}(\tilde{\theta})(\hat{\theta}_N - \theta_0). \\ In the last line, we use the fact that the expected value of the score function (derivative of log likelihood) is zero. \end{aligned} \tag{19} \\ The second case seems to follow CLT well, however both cases are already proved and easily seen anywhere. \tag{29} LN()=N1(22logfX(X;))=N1(22logn=1NfX(Xn;))=N1n=1N(22logfX(Xn;))pE[22logfX(X1;)].(12), In the last step, we invoke the WLLN without loss of generality on X1X_1X1. Examples include: (1) bN is an estimator, say b;(2)bN is a component of an estimator, such as N1 P ixiui;(3)bNis a test statistic. /Length 2144 /Subtype/Form LN()LN()LN()=N1logfX(x;),=(N1logfX(x;)),=22(N1logfX(x;)).(3). I thank Sam Morin for pointing out a couple mistakes in the Bernoulli derivations. With these properties in mind, lets prove some important facts about the OLS estimator ^\hat{\boldsymbol{\beta}}^. HpE$L@#: (19) (26) ") K8fuA,%dN4*AyPQ[}F)8[&4O@=& hFT$)2ji~2vf$cdzbn$|qHd,V7=[8#*d\dZHwIP&"j} Lets write out the vectorization explicitly, E[MX]=E[[1N][M11M1NMN1MNN][1N]|X]=E[[1N][i=1NM1iii=1NMNii]|X]=E[j=1Ni=1NMjiji|X](20) 0000004083 00000 n Wbp4z6%Cq6XF\_mV1*IPTQKTPcS>7d|B3U" Q"mG]CG_P!R~%+0,KJXHB%$wvwz8 mU}U'm_'Nd6*v.uV5Vys^`;k-5"gqT!HRddZHwIP2$AF -i53283.Iax.5/mBR N(^N0)dN(I(0)1). 0000001376 00000 n The proof is, M=(yX)M(yX)=yMy+XMXyMXXMy=yMy=yMMy=ee. The proof of the main theorem appears in the appendix. (16) LN(~)pI(0).(15). Modified 4 years, 9 months ago. plogfX(X;p)=n=1N[pXnlogp+p(1Xn)log(1p)]=n=1N[pXn1p1Xn]=n=1N[pXn+1pXn1].(19). \mathbb{E}[\varepsilon_n \varepsilon_m \mid \mathbf{X}] = 0, \quad n,m \in \{1, \dots, N\}, \quad n \neq m. \tag{4} \end{aligned} \tag{17} Theorem 11 (On asymptotic normality)Suppose the assumptionsH1-H5hold, and in addition it is assumed that: 1 the fourth moment is finite, i.e., 2 the matrix is stable14. \begin{bmatrix} \varepsilon_1 & \dots & \varepsilon_N \end{bmatrix} p^N=N1n=1NXn. }=+UaHD2 In many econometric situations, normality is not a realistic assumption \tag{24} 0000003584 00000 n \begin{aligned} Since this is a quadratic expression, the vector which gives the global minimum may be found via matrix calculus by differentiating with respect to the vector (using denominator layout) and setting equal to zero: By assumption matrix X has . fZ=Rr&P*EDqL%)P&b!o2HWF6ABDbVJh Asymptotics Part IVb The delta method M- and Z-Estimators . %PDF-1.4 % Let X1,,XNX_1, \dots, X_NX1,,XN be i.i.d. &= N - \text{trace}\left( \mathbf{I}_P \right) 0000029003 00000 n \mathcal{I}_N(p) Now lets apply the mean value theorem, Mean value theorem: Let fff be a continuous function on the closed interval [a,b][a, b][a,b] and differentiable on the open interval. 0000017196 00000 n /Length 23 0000049478 00000 n \begin{aligned} To perform tasks such as hypothesis testing for a given estimated coefficient ^p\hat{\beta}_p^p, we need to pin down the sampling distribution of the OLS estimator ^=[1,,P]\hat{\boldsymbol{\beta}} = [\beta_1, \dots, \beta_P]^{\top}^=[1,,P]. Proof: We shall derive it as a special case when doing GMM testing . The negative sign is due to the chain rule when computing the derivative of log(1p)\log(1-p)log(1p). Assumption 1 is just Equation 1; it means that we have correctly specified our model. If asymptotic normality holds, then asymptotic efficiency falls out because it immediately implies, ^NdN(0,IN(0)1). Knowing this distribution is useful in analyzing the results of linear models, such as when performing hypothesis testing for a given estimated parameter ^p\hat{\beta}_p^p. &\stackrel{\ddagger}{=} \mathbf{e}^{\top} \mathbf{e}. trace(M)=trace(INH)=trace(IN)trace(H)=Ntrace(X(XX)1X)=Ntrace(XX(XX)1)=Ntrace(IP)=NP.(24), If we make assumption 555, that the error terms are normally distributed, then ^\hat{\boldsymbol{\beta}}^ is also normally distributed. The terms (1p)(1 - p)(1p) cancel, leaving us with the MLE: p^N=1Nn=1NXn. Step \star is because the true value \boldsymbol{\beta} is non-random; step \dagger is just applying Equation 555 from above; step \ddagger is because A\mathbf{A}A is non-random; and step * is assumption 444 or spherical errors. E[^X]=. %PDF-1.5 \\ # Generate many random samples of size N and compute MLE. Assumption 444 can be broken into two assumptions. &\stackrel{\dagger}{=} \mathbb{V}[\overbrace{(\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top}}^{\mathbf{A}} \boldsymbol{\varepsilon} \mid \mathbf{X}] \tag{26} \sqrt{N} L^{\prime}_N(\theta_0) \rightarrow^d \mathcal{N}\left(0, \mathbb{V}\left[\frac{\partial}{\partial \theta} \log f_X(X_1; \theta_0)\right]\right). The term in the expectation, ^\hat{\boldsymbol{\beta}} - \boldsymbol{\beta}^, is sometimes called the sampling error, and we can write it in terms of the predictors and noise terms: ^=(XX)1Xy=(XX)1X(X+)=(XX)1(XX)+(XX)1X)=+(XX)1X=(XX)1X. 0000028224 00000 n OLS Asymptotics PaulSchrimpf Motivation Consistency Asymptotic normality Largesample inference References Reviewofcentrallimittheorem LetFn betheCDFof andWbearandomvariablewith CDFF convergesindistributiontoW,written d W,if limn Fn(x) = F(x) forallxwhereFiscontinuous Centrallimittheorem:Let{y1,.,yn} bei.i.d.with mean andvariance2 thenZn = we run a OLS regression of lnw i on a constant and educ i; b ols = 1 n P n i=1 lnw i 1 n P n i=1 lnw i educ i 1 P n . Dr. Henry Asymptotic normality of OLS Kankwamba. \middle| \mathbf{X} \right] Difference between asymptotic normalities of OLS and MLE. The left cancellation is because MX=0\mathbf{M}\mathbf{X} = \mathbf{0}MX=0: MX=(INH)X=XHX=XX(XX)1XX=0. (6), Above, we have just rearranged terms. /Filter/FlateDecode So we can move the MjiM_{ji}Mji terms out of the expectation to get, E[M]=j=1Ni=1NMjiE[ji|X]. - We come to this approximation by the CLT because the OLS estimators involve the use of sample averages (mathematically, this can get complicated). \tag{15} 0000001857 00000 n Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Asymptotic Normality Large Sample Inference t, F tests based on normality of the errors (MLR.6) if drawn from other distributions j will not be normal t, F statistics will not have t, F distributions solutionuse CLT: OLS estimators are approximately normally distributed (at least for large sample sizes) \mathbf{E}[\boldsymbol{\varepsilon}^{\top} \mathbf{M} \boldsymbol{\varepsilon}] = \sum_{j=1}^N \sum_{i=1}^N M_{ji} \mathbb{E} \left[ \varepsilon_j \varepsilon_i \middle| \mathbf{X} \right]. (6), E[^X]=0. \begin{aligned} >> &= -\mathbb{E}\left[ \sum_{n=1}^N \left[ - \frac{X_n}{p^2} + \frac{X_n - 1}{(1 - p)^2} \right] \right] (17) 0000017450 00000 n x\Ks8Q>6y60JcZ#)3? 0000028009 00000 n \begin{aligned} /FormType 1 Using basic properties of the normal distribution, we can immediately derive the distribution of the OLS estimator: ^N(,2(XX)1). \end{aligned} \tag{20} Finally, a set of simulations illustrate the asymptotic behavior of the OLS. where M\mathbf{M}M is the residual maker, M=X(XX)1X,(14) \mathbb{V}\left[\frac{\partial}{\partial \theta} \log f_X(X_1; \theta_0)\right] \frac{\partial}{\partial p} \log f_X(X; p) 925 0 obj <>stream NLN(0)dN(0,I(0))(14), LN(~)pI(0). Now note that ~(^N,0)\tilde{\theta} \in (\hat{\theta}_N, \theta_0)~(^N,0) by construction, and we assume that ^Np0\hat{\theta}_N \rightarrow^p \theta_0^Np0. \tag{17} %PDF-1.5 % hbbd``b`A`@Y (16) Theorem 2 is a special case of a central limit theorem (holding uniformly on [Math Processing Error] F ( F)) for families of random sequences [3]. There are no real details on the CLT, j. (b6=!lI~u^ ?)>]W B-3)%dH dG%[)c_JL%r3!Z9KfzDH3YDEM955)w4#tN sZg~{1gCto .L?| \end{aligned} \tag{3} Given a statistical model P\mathbb{P}_{\theta}P and a random variable XP0X \sim \mathbb{P}_{\theta_0}XP0 where 0\theta_00 are the true generative parameters, maximum likelihood estimation (MLE) finds a point estimate ^N\hat{\theta}_N^N such that the resulting distribution most likely generated the data. Then we can invoke Slutskys theorem. \end{aligned} \tag{12} 0 The OLS coefficient estimator 1 is unbiased, meaning that . &= \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1}. \end{aligned} \tag{12} 0000014503 00000 n /Subtype/Form Nonetheless, it is relatively easy to analyze the asymptotic performance of the OLS estimator and construct large-sample tests. trace(M)=NP.(16). that, E[^X]=. To prove that s2s^2s2 is unbiased, it suffices to show that, E[s2X]=2,E[eeX]=(NP)2. (18) \\ /Length 23 0000049280 00000 n 0 yn=0+1xn,1++Pxn,P+n.(1). \mathbb{E}[\varepsilon_n \mid \mathbf{X}] = 0, \quad n \in \{1, \dots, N\}. Here is a nice example of why Equation 222 captures this intuition. stream E[M]=iNMii2=trace(M)2.(23). \hat{\theta}_N = \arg\!\max_{\theta \in \Theta} \log f_X(x; \theta) \quad \implies \quad L^{\prime}_N(\hat{\theta}_N) = 0. According to the asymptotic properties of the OLS estimator: OLS is consistent, The estimator converges in distribution to standard normal, Inference can be performed based on the asymptotic convergence to the standard normal, and OLS is the most efficient among many consistent estimators of .

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