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Thread: probability denstity function of estimator . thanks:)

  1. #1
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    probability denstity function of estimator . thanks:)

    I have random variable X with uniform distribution on interval [0, B], thus it has pdf f(x)=1/B when 0<=x<=B and 0 otherwise. Then its dumulative distribution function is x/B, 1 for x>B and 0 if X<0. For the given uniform distribution the E(x)= m= B/2 and Var(x)= s^2 = B^(2)/12.
    If i.i.d is taken from this distribution x1, x2,....xn. and for the estimation of the expected value m=B/2 we have defined estimotors m1= x bar the sample average and m2= M/2, where M is the sample maximum the largest element of the sample.

    MY PROBLEM:

    To show that the sampling distribution of M has the following pdf.

    f(m)=(n/B) *(m/B)^(n-1)

    I assume the way to start is with the cumulative distribution function
    F(m)= Pr(m<M) =.... {DO NOT KNOW HOW TO GO ON....}

    Moreover i am not sure whether i got it right for the E (m2) and the Var(m2)

    Is it the E(m) = B/4.


    Any help is welcome!

    Many thanks
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  2. #2
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    Quote Originally Posted by saskadimova View Post
    I have random variable X with uniform distribution on interval [0, B], thus it has pdf f(x)=1/B when 0<=x<=B and 0 otherwise. Then its dumulative distribution function is x/B, 1 for x>B and 0 if X<0. For the given uniform distribution the E(x)= m= B/2 and Var(x)= s^2 = B^(2)/12.
    If i.i.d is taken from this distribution x1, x2,....xn. and for the estimation of the expected value m=B/2 we have defined estimotors m1= x bar the sample average and m2= M/2, where M is the sample maximum the largest element of the sample.

    MY PROBLEM:

    To show that the sampling distribution of M has the following pdf.

    f(m)=(n/B) *(m/B)^(n-1)

    I assume the way to start is with the cumulative distribution function
    F(m)= Pr(m<M) =.... {DO NOT KNOW HOW TO GO ON....}

    Moreover i am not sure whether i got it right for the E (m2) and the Var(m2)

    Is it the E(m) = B/4.


    Any help is welcome!

    Many thanks
    Let $\displaystyle X_1$, $\displaystyle X_2$, .... $\displaystyle X_n$ denote independent continuous random variables with distribution function $\displaystyle F(x)$ and density function $\displaystyle f(x)$. Let the ordered random variables $\displaystyle X_i$ be denoted by $\displaystyle X_{(1)}$, $\displaystyle X_{(2)}$, .... $\displaystyle X_{(n)}$ where $\displaystyle X_{(1)} \leq X_{(2)} \leq .... X_{(n)}$. That is, $\displaystyle X_{(1)} = \text{min}(X_1, .... X_n)$ and$\displaystyle X_{(n)} = \text{max}(X_1, .... X_n)$.

    The probability density function for $\displaystyle X_{(n)}$ can be found as follows:

    Since $\displaystyle X_{(n)}$ is the maximum of $\displaystyle X_1$, $\displaystyle X_2$, .... $\displaystyle X_n$, then:

    $\displaystyle G_n(x) = \Pr(X_{(n)} \leq x) = \Pr(X_1 \leq x, X_2 \leq x, .... X_n \leq x)$ $\displaystyle = \Pr(X_{(1)} \leq x) \cdot \Pr(X_{(2)} \leq x) \cdot .... \cdot \Pr(X_{(n)} \leq x) = [F(x)]^n$.

    Let $\displaystyle g_n(x)$ denote the pdf of $\displaystyle X_{(n)}$. Then $\displaystyle g_n(x) = \frac{dG_n(x)}{dx} = n [F(x)]^{n-1} f(x)$.

    Substitute the pdf and cdf of your distribution into the above formula and you will have your answer.


    Note: This stuff is known as Order Statistics.
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  3. #3
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    What about the expected value of the second estimotr m2.

    Is it B/4 and therfore biased or i am making mistakes?

    thx!

    Btw is any specific place on the forum dedicated to order statistics?
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  4. #4
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    Quote Originally Posted by saskadimova View Post
    What about the expected value of the second estimotr m2. Mr F says: What about it? If you have the pdf for it you should be able to calculate its expected value ....

    Is it B/4 and therfore biased or i am making mistakes? Mr F says: I don't know. I haven't done it. But if you show all your work I'll review it.

    thx!

    Btw is any specific place on the forum dedicated to order statistics? Mr F says: Questions about order statistics can be posted in this subforum.
    ..
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