Let us give one more example of MLE. Example 3. The uniform distribution U[0, θ] on the interval [0, θ] has p.d.f.
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1 Lecture 5 Let us give oe more example of MLE. Example 3. The uiform distributio U[0, ] o the iterval [0, ] has p.d.f. { 1 f(x =, 0 x, 0, otherwise The likelihood fuctio ϕ( = f(x i = 1 I(X 1,..., X [0, ] = 1 I(max(X 1,..., X. Here the idicator fuctio I(A equals to 1 if A happes ad 0 otherwise. What we wrote is that the product of p.d.f. f(x i will be equal to 0 if at least oe of the factors is 0 ad this will happe if at least oe of X i s will fall outside of the iterval [0, ] which is the same as the maximum amog them exceeds. I other words, ad ϕ( = 0 if < max(x 1,..., X, ϕ( = 1 if max(x 1,..., X. Therefore, lookig at the figure 5.1 we see that ˆ = max(x 1,..., X is the MLE. 5.1 Cosistecy of MLE. Why the MLE ˆ coverges to the ukow parameter 0? This is ot immediately obvious ad i this sectio we will give a sketch of why this happes. 17
2 LECTURE ϕ( max(x1,..., X Figure 5.1: Maximize over First of all, MLE ˆ is a maximizer of L = 1 log f(x i which is just a log-likelihood fuctio ormalized by 1 (of course, this does ot affect the maximizatio. L ( depeds o data. Let us cosider a fuctio l(x = log f(x ad defie L( = 0 l(x, where we recall that 0 is the true ukow parameter of the sample X 1,..., X. By the law of large umbers, for ay, L ( 0 l(x = L(. Note that L( does ot deped o the sample, it oly depeds o. We will eed the followig Lemma. We have, for ay, L( L( 0. Moreover, the iequality is strict L( < L( 0 uless which meas that = 0. 0 (f(x = f(x 0 = 1.
3 LECTURE Proof. Let us cosider the differece L( L( 0 = 0 (log f(x log f(x 0 = 0 log f(x f(x 0. t 1 log t t 0 1 Figure 5.2: Diagram (t 1 vs. log t Sice (t 1 is a upper boud o log t (see figure 5.2 we ca write 0 log f(x f(x 0 = ( f(x ( f(x 0 f(x 0 1 = f(x 0 1 f(x 0 dx f(x dx f(x 0 dx = 1 1 = 0. Both itegrals are equal to 1 because we are itegratig the probability desity fuctios. This proves that L( L( 0 0. The secod statemet of Lemma is also clear. We will use this Lemma to sketch the cosistecy of the MLE. Theorem: Uder some regularity coditios o the family of distributios, MLE ˆ is cosistet, i.e. ˆ 0 as. The statemet of this Theorem is ot very precise but but rather tha provig a rigorous mathematical statemet our goal here to illustrate the mai idea. Mathematically iclied studets are welcome to come up with some precise statemet. Proof.
4 LECTURE We have the followig facts: 1. ˆ is the maximizer of L ( (by defiitio is the maximizer of L( (by Lemma. 3. we have L ( L( by LLN. This situatio is illustrated i figure 5.3. Therefore, sice two fuctios L ad L are gettig closer, the poits of maximum should also get closer which exactly meas that ˆ 0. L( L( ^ MLE 0 Figure 5.3: Lemma: L( L( Asymptotic ormality of MLE. Fisher iformatio. We wat to show the asymptotic ormality of MLE, i.e. that (ˆ 0 d N(0, σ 2 MLE for some σ2 MLE. Let us recall that above we defied the fuctio l(x = log f(x. To simplify the otatios we will deote by l (X, l (X, etc. the derivatives of l(x with respect to.
5 LECTURE Defiitio. (Fisher iformatio. Fisher Iformatio of a radom variable X with distributio 0 from the family { : Θ} is defied by I( 0 = 0 (l (X ( log f(x 0 2. Next lemma gives aother ofte coveiet way to compute Fisher iformatio. Lemma. We have, ad Proof. First of all, we have 0 l (X Also, sice p.d.f. itegrates to 1, 2 log f(x 0 = I( 0. l (X = (log f(x = f (X f(x (log f(x = f (X f(x (f (X 2 f 2 (X. f(x dx = 1, if we take derivatives of this equatio with respect to (ad iterchage derivative ad itegral, which ca usually be doe we will get, f(x dx = 0 ad 2 f(x dx = 2 f (x dx = 0. To fiish the proof we write the followig computatio 0 l 2 (X 0 = 0 log f(x 0 = (log f(x 2 0 f(x 0 dx (f (x 0 ( f = f(x 0 (x 0 2 f(x 0 dx f(x 0 = f (x 0 dx 0 (l (X 0 2 = 0 I( 0 = I( 0. We are ow ready to prove the mai result of this sectio.
6 LECTURE Theorem. (Asymptotic ormality of MLE. We have, 1 (ˆ 0 N( 0,. I( 0 Proof. Sice MLE ˆ is maximizer of L ( = 1 log f(x i we have, Let us use the Mea Value Theorem f(a f(b a b L (ˆ = 0. = f (c or f(a = f(b + f (c(a b for c [a, b] with f( = L (, a = ˆ ad b = 0. The we ca write, 0 = L (ˆ = L ( 0 + L (ˆ 1 (ˆ 0 for some ˆ 1 [ˆ, 0 ]. From here we get that ˆ 0 = L ( 0 L (ˆ 1 ad L (ˆ 0 = ( 0 L (ˆ. (5.1 1 Sice by Lemma i the previous sectio 0 is the maximizer of L(, we have Therefore, the umerator i (5.1 L ( 0 = ( 1 = ( 1 L ( 0 = 0 l (X 0 = 0. (5.2 l (X i 0 0 ( l (X i 0 0 l (X 1 0 N 0, Var 0 (l (X 1 0 (5.3 coverges i distributio by Cetral Limit Theorem. Next, let us cosider the deomiator i (5.1. First of all, we have that for all, L ( = 1 l (X i 0 l (X 1 by LLN. (5.4 Also, sice ˆ 1 [ˆ, 0 ] ad by cosistecy result of previous sectio ˆ 0, we have ˆ 1 0. Usig this together with (5.4 we get L (ˆ 1 0 l (X 1 0 = I( 0 by Lemma above.
7 LECTURE Combiig this with (5.3 we get L ( 0 L (ˆ 1 Fially, the variace, ( N 0, Var 0 (l (X 1 0. (I( 0 2 Var 0 (l (X 1 0 = 0 (l (X 0 2 ( 0 l (x 0 2 = I( 0 0 where i the last equality we used the defiitio of Fisher iformatio ad (5.2.
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