One obervation is taken from the probability model. f(x;θ)=((θ/2)^|x|)*((1-θ)^(1-|x|)) for x=-1,0,1 and θ ∈ [0,1]. We...

One obervation is taken from the probability model. f(x;θ)=((θ/2)^|x|)*((1-θ)^(1-|x|)) for x=-1,0,1 and θ ∈ [0,1]. We find that the MLE of θ is |x|.

Now consider T(X) where T(X)=2 when x=1 and T(X)=0 otherwise. Show that the MLE of θ has a smaller variance than T(X).

Expert Solution

orchestra answered 1 year ago

We have a one parameter statistical model. Probability model: {f(x;θ)=x/(2θ^2), θ>0 and x is an element...

We have a one parameter statistical model. Probability model: {f(x;θ)=x/(2θ^2), θ>0 and x is an element of (0, θ]. Sampling model: X=(X1,...Xn) is random. By stating likelihood and log likelihood functions, calculate the maximum likelihood estimator of θ. Differentiation can't be used to solve this problem.

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Let X1. . . . Xn be i.i.d f(x; θ) = θ(1 − θ)^x x =...

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Let X1,X2,...,Xn be i.i.d. (independent and identically distributed) from the Bernoulli distribution f(x)=p^x(1-p)^1-x, x=0,1,p∈(0,1) where p...

Let X1,X2,...,Xn be i.i.d. (independent and identically distributed) from the Bernoulli distribution f(x)=p^x(1-p)^1-x, x=0,1,p∈(0,1) where p is unknown parameter. Find the UMVUE of p parameter and calculate MSE (Mean Square Error) of this UMVUE estimator.

If we rename y=x^2, f(x)=x^2, could we rename the inverse f^-1(x)? Explain your answer

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