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statistics question statistics question

01-31-2012 , 01:03 PM
hello, does anyone know how to solve the following problem?

E(X)=2a/3 V(X)=unknown now you have to check the medium squared error of the assessor: â=Σ Xi . As we dont know Var(X) and I am not sure how X is distributed I dont know how I can get Var(X)?

f(x)= (2/a²)x for "0 <= x <= a" otherwise "0"
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01-31-2012 , 01:19 PM
f(x) is the pdf for x. http://en.wikipedia.org/wiki/Variance#Definition I have no idea what the assessor part is if you don't have some kind of sample data points given. Also, homework thread.
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01-31-2012 , 01:32 PM
Quote:
Originally Posted by TomCowley
f(x) is the pdf for x. http://en.wikipedia.org/wiki/Variance#Definition I have no idea what the assessor part is if you don't have some kind of sample data points given. Also, homework thread.
I think I can solve it using the integral formula, have to try that one but looks good.

is there a homework thread?
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01-31-2012 , 01:47 PM
Quote:
Originally Posted by TomCowley
f(x) is the pdf for x. http://en.wikipedia.org/wiki/Variance#Definition I have no idea what the assessor part is if you don't have some kind of sample data points given. Also, homework thread.
sry for my bad english, first of all didnt get nowhere unfortunately. Second I didnt mean to say assessor, I meant estimator = "â= Σ Xi"
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01-31-2012 , 04:31 PM
Use VAR(X)=E(X²)-E(X)², to compute the variance, it's often easier. Especially if you're given E(X). Now you only need E(X²). As for that estimator... sure you're not forgetting to divide by the number of observations?
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01-31-2012 , 05:25 PM
Quote:
Originally Posted by Vael
Use VAR(X)=E(X²)-E(X)², to compute the variance, it's often easier. Especially if you're given E(X). Now you only need E(X²). As for that estimator... sure you're not forgetting to divide by the number of observations?
That was my approach too. using the Var(x)=E(x^2)-E(x)^2 formula. The problem is that I neither know the value vor Var(x) nor E(x^2), so its like I have a formula with two unknowns...(unless I miss something).

if I use the formula I get VAR(X)=E(X^2)-n*(2a/3)^2 and there it stops.

so I dont really understand what you mean by dividing through the number of observations.
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01-31-2012 , 06:51 PM
from the link TC posted:


That's how you derive E(X) from the probability density function you're given. You don't need that, because you know E(X) already . What you need is E(X²), so integrate x²f(x) instead of xf(x).

Well, the Xi are observations, produced by the pdf you're given, right? If you take your â and divide by n, the expected value of that expression is 2a/3. Rearranging gives you an estimator for the parameter a in terms of your observations (namely the sample mean: your â divided by n)
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01-31-2012 , 07:11 PM
Quote:
Originally Posted by Vael
from the link TC posted:


That's how you derive E(X) from the probability density function you're given. You don't need that, because you know E(X) already . What you need is E(X²), so integrate x²f(x) instead of xf(x).

Well, the Xi are observations, produced by the pdf you're given, right? If you take your â and divide by n, the expected value of that expression is 2a/3. Rearranging gives you an estimator for the parameter a in terms of your observations (namely the sample mean: your â divided by n)
Now I got!!! thx very much, after learning for eight hours straight you cant put some things together you would probably realize if fit^^
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