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Thread: MLE of P(X<2) - Exponential distribution

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    MLE of P(X<2) - Exponential distribution



    The problem statement:
    Find the MLE of θ = P (X≤ 2) in a random sample of size n selected from an exponential distribution EXP(λ)

    Relevant equations

    f(x, λ) = λ e^(-λx)
    F(x, λ) = 1 - e^(-λx)

    The attempt at a solution
    I know how to find the MLE of the mean of an exponential distribution. But I am not sure how I can tackle this problem.

    We know that P ( X≤ 2) = ∫f(x) 0,2 = F(4)

    How do I get to the Likelihood from here?

    Thanks!

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    Re: MLE of P(X<2) - Exponential distribution

    If \theta = \Pr\{X \leq 2\} = g(\lambda), then by the functional invariance of MLE, the MLE of \theta is \hat{\theta} = g(\hat{\lambda}) where \hat{\lambda} is the MLE of \lambda. This is particularly easy to see if g is a bijective function.

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    sandeep249 (04-21-2012)

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    Re: MLE of P(X<2) - Exponential distribution


    Thanks, that was very helpful!

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