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Thread: Exponential law exercise

  1. #1
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    Exponential law exercise

    Hello ! I am struggling a bit with this probability exercise. Thank you very much for any help with the problem !


    X is a continuous random variable distributed following an exponential law of parameter z with a density
    Fx(x) = z*exp(-z*x) when x>= 0
    Fx(x) = 0 when x<0

    1) Write the expression of the distribution function of X and give the probability that X goes above the value x= 1 when z = 1



    I found it by searching on wikipedia, however I don't know how to find it from the density. Was I supposed to derive this expression ?


    F(x) = P(X <= 1)
    F(x) = 1 - exp(-1*1) = 1 - exp(-1)

    So for X’s value to go above 1 : 1 – (1 - exp(-1))


    2) Using the maximum likelihood method, give the expression of the estimator of z

    I do not know at all how to proceed for this question
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  2. #2
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    Re: Exponential law exercise

    1) you've done correctly.

    2) basically you want to find, given an observation $x$, the parameter $z$ which is the most likely to have caused the observation.

    we find the maximum $z$ the way we find any other maximum of a function, setting it's derivative to zero and solving.

    $f_z(x) = z e^{-z x}$

    $\dfrac{\partial f}{\partial z} = e^{-zx} - z x e^{-z x} = (1-x z) e^{-zx}$

    $f_z(x) = 0 \Rightarrow z = \dfrac 1 x$

    This isn't surprising since $E[x] = \dfrac 1 z$
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  3. #3
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    Re: Exponential law exercise

    Thank you for the explanation !

    When I try to derive fx I obtain -z^2 e^(-zx) though. Is there a différence between df/dz and df/dx ?
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  4. #4
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    Re: Exponential law exercise

    Quote Originally Posted by Jo37 View Post
    Thank you for the explanation !

    When I try to derive fx I obtain -z^2 e^(-zx) though. Is there a différence between df/dz and df/dx ?
    you really should know the answer to that one.

    of course there is.

    you are trying to maximize the density function by adjusting $z$ not $x$. You have a value for $x$, that's your observation.

    so you use $\dfrac{df}{dz}=0$
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