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    Consistent MLE of a U(0, θ)




    So the question is:
    Let X1,.., Xn ∼ U(0, θ). Show that the MLE is consistent.
    Hint: Let Y=max {X1,... ,Xn}, for any c, P(Y <c)=P(X1 <c,X2 <c,...,Xn <c)=P(X1 <c)P(X2 < c)...P(Xn < c).
    So I know for the MLE to be consistent, the estimated value of theta has to converge (in probability) to the actual value.
    Im trying to think of how this hint given can lead me to the result.

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    Re: Consistent MLE of a U(0, θ)

    The hint is quite good. If you don't see how that helps then try plugging in some values - ie set a value for c, assume n = 2 and calculate P(Y < c). Now give it a try for n = 3, n = 4, ... you should something happening and hopefully you can figure out why it is happening.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Consistent MLE of a U(0, θ)

    So because its a uniform distributions won't all the P(X_{i}<c)=c/θ?
    So P(Y<c)=(c/θ)^n?

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    Re: Consistent MLE of a U(0, θ)

    Yes. And what happens to that as n gets large?
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    Re: Consistent MLE of a U(0, θ)

    Well, it goes to zero, but how does that show that the MLE is consistent?

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    Re: Consistent MLE of a U(0, θ)

    What do you need to be true for the MLE to be consistent? You wrote out the condition in the first post but write out mathematically what you need to be true for us.
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    Re: Consistent MLE of a U(0, θ)

    Well I said that the Estimator has to converge to the actual value, which we found is zero.
    So if MLE is \hat\theta=max(X_{1}+...+X_{n}) this has to converge to 0 in probability.
    Which we just showed using the hint?

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    Re: Consistent MLE of a U(0, θ)

    Quote Originally Posted by calculuskid1 View Post
    Well I said that the Estimator has to converge to the actual value, which we found is zero.
    0? You sure about that? What we showed is that for any c < \theta that \lim_{n \rightarrow \infty}P(Y_n < c) = 0.

    Do you really think the max of your sample should converge to 0?

    And I was asking you to write out mathematically what it is that we need to show to say that a random variable converges in probability.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Consistent MLE of a U(0, θ)

    Well obviously it does not make sense for the max to converge to zero...
    So we have to show that:
    lim P(|\theta - \hat\theta| >= \epsilon) = 0
    ?

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    Re: Consistent MLE of a U(0, θ)

    I am looking at examples online and it seems like they are all taking the expected value of the difference of the two.
    So E(\hat\theta - \theta)^2=2\theta^2/(n+1)(n+2)

    And this obviously goes to 0 and n is large but Im not sure where the (n+1)(n+2) comes from

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    Re: Consistent MLE of a U(0, θ)

    Notice that \hat{\theta} = Y and that we can rewrite |\theta - Y| \geq \epsilon as Y \leq \theta - \epsilon (using the fact that Y \leq \theta)

    Can you see how to apply the result from earlier now?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Consistent MLE of a U(0, θ)


    Yes I see, basically let c=\theta - \epsilon and the rest we have done lol

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