MLE of P(X<2) - Exponential distribution

SandMan249
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Homework Statement



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

Homework 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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SandMan249 said:

Homework Statement



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

Homework 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!

Is the following statement of your problem correct? You observe n independent values of X and observe that m of them have {X < 2} (or {X <= 2}). From that, you want to estimate θ = P{X <= 2}. If that is truly the statement, what does the exponential distribution have to do with it? (Of course, if you want to estimate λ you need to know the distribution, but that is not what you said you want to estimate.)

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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