SOLUTION FOR HOMEWORK 3, STAT 4352 Welcome to your third ...

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This is the answer. But again, as an extra example, I can suggest a MME based on the second moment. Indeed, Eλ(X 2 ) = V arλ(X) + (EλX) 2 = λ + λ 2 and this yields that ˜λMME + ˜ λ 2 n MME = n−1 X i=1 2 i . Then you need to solve this equation to get the MME. Obviously it is a more complicated estimator, but it is yet another MME. 3. Problem 10.56. Let X1, . . .,Xn be iid according to the pdf gθ(x) = θ −1 e −(x−δ)/θ I(x > δ). Please note that this is a location-exponential family because X = δ + Z, where Z is a classical exponential RV with f Z θ (z) = θ −1 e −z/θ I(z > 0). I can go either further by saying that we are dealing with a location-scale family because X = δ + θZ0, where f Z0 (z) = e −z I(z > 0). So now we know the meaning of parameters δ and θ: the former is the location (shift) and the latter is the scale (multiplier). Note that this understanding simplifies all calculations because you can easily figure out (otherwise do calculations) that Eδ,θ(X) = δ + θ, V arδ,θ(X) = θ 2 . These two familiar results yield Eδ,θ(X 2 ) = θ 2 + (δ + θ) 2 , and we get the following system of two equations to find the pair of MMEs: ˆδ + ˆ θ = ¯ X, 2ˆ θ 2 + 2ˆ δˆ θ + ˆ δ 2 = n −1 n Xi. i=1 To solve this system, we square the both sides of the first equality and then subtract the obtained equality from the second equality. We get a new system ˆδ + ˆ θ = ¯ X, ˆθ 2 = n −1 n X i=1 2 i − ¯ X 2 . This, together with a simple algebra, yields the answer ˆδMME = ¯ X − [n −1 n i=1 X 2 i − ¯ X 2 ] 1/2 , θMME ˆ = [n −1 n X 2 i − ¯ X 2 ] 1/2 . 2 i=1
Remark: We need to check that n −1 n i=1 X 2 i − ¯ X 2 ≥ 0 for the estimator to be well defined. This may be done via famous Hölder inequality m ( aj) j=1 2 m ≤ m a j=1 2 j . 4. Problem 10.59. Here X1, . . ., Xn are Poisson(λ), λ ∈ Ω = (0, ∞). Recall that Eλ(X) = λ and V arλ(X) = λ. Then, by definition of the MLE: ˆλMLE := arg max λ∈Ω = arg max λ∈Ω n l=1 n l=1 For the Poisson pdf fλ(x) = e −λ λ x /x! we get ln LXn(λ) = −nλ + fλ(Xl) =: arg max LXn(λ) λ∈Ω ln(fλ(Xl)) =: arg max ln LXn(λ). λ∈Ω n n Xl ln(λ) − ln(Xl!). l=1 l=1 Now we need to find ˆ λMLE at which the above loglikelihood attains its maximum over all λ ∈ Ω. You can do this in a usual way: take derivative with respect to λ ( that is, calculate ∂ lnLX n(λ)/∂λ, then equate it to zero, solve with respect to λ, and then check that the solution indeed maximizes the loglikelihood). Here equating of the derivative to zero yields −n + n l=1 Xl/λ = 0, and we get ˆλMLE = ¯ X. Note that for the Poisson setting the MME and MLE coincide; in general they may be different. 5. Problem 10.62. Here X1, . . .,Xn are iid N(µ, σ 2 ) with the mean µ being known and the parameter of interest being the variance σ 2 . Note that σ 2 ∈ Ω = (0, ∞). Then we are interested in the MLE. Write: Here ˆσ 2 MLE = arg max σ 2 ∈Ω ln LX n(σ2 ). ln LXx(σ2 n ) = ln([2πσ l=1 2 ] −1/2 e −(Xl−µ) 2 /(2σ2 n ) 2 2 ) = −(n/2) ln(2πσ ) − (1/2σ ) (Xl − µ) l=1 2 . This expression takes on its maximum at Note that this is also the MME. ˆσ 2 MLE = n−1 n (Xl − µ) 2 . l=1 3
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Page 5 and 6: extreme observations is also the mi
Page 7 and 8: which in its turn yields Using this
Page 9: 12. Problem 10.96. Let X be a grade

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