# Math Help - probability denstity function of estimator . thanks:)

1. ## 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

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

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

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

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

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

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.

3. 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?