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thesandbox
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Homework Statement
If you were taking a random sample of size n (n=2m+1 odd) from Uni(0,1)
How do you find the mean and variance of the sample median?
Homework Equations
In order to find the mean and variance of the sample median you need to start with the sample median itself. Using this equation:
ƒxmedian(x) = ƒx(2m+1)(x) = [itex]\frac{(2m+1)!}{m!m!}[/itex]*ƒ(x)*[F(x)]m*[1-F(x)]m
Where ƒ(x) is the pdf of the Uni(0,1) ~ Uni(a,b)
ƒ(x) = [itex]\frac{1}{b-a}[/itex] This becomes = 1
Where F(x) is the cdf of the Uni(0,1)
F(x) = [itex]\frac{x-a}{b-a}[/itex] This becomes x
So,
ƒxmedian(x)
= ƒx(2m+1)(x)
= [itex]\frac{(2m+1)!}{m!m!}[/itex]*(1)*xm*(1-x)m
= [itex]\frac{(2m+1)!}{m!m!}[/itex]*xm*(1-x)m
The Attempt at a Solution
Above is part of the attempt.
Now as for the mean and variance of the sample median
Mean
E(x) = x*ƒxmedian(x)*dx
= [itex]\int[/itex][itex]^{1}_{0}[/itex] x*[itex]\frac{(2m+1)!}{m!m!}[/itex]*xm*(1-x)m
= [itex]\frac{(2m+1)!}{m!m!}[/itex]*[itex]\int[/itex][itex]^{1}_{0}[/itex] x*xm*(1-x)m
= [itex]\frac{(2m+1)!}{m!m!}[/itex]*[itex]\int[/itex][itex]^{1}_{0}[/itex] xm+1*(1-x)m
= [itex]\frac{(2m+1)!}{m!m!}[/itex]*[itex]\int[/itex][itex]^{1}_{0}[/itex] x*xm*(1-x)m
= [itex]\frac{(2m+1)!}{m!m!}[/itex]*[itex]\int[/itex][itex]^{1}_{0}[/itex] x*(xm*(1-x)m)
= [itex]\frac{(2m+1)!}{m!m!}[/itex]*[itex]\int[/itex][itex]^{1}_{0}[/itex] x*[x*(1-x)]m
= [itex]\frac{(2m+1)!}{m!m!}[/itex]*[itex]\int[/itex][itex]^{1}_{0}[/itex] x*[(x-x2)]m
-> Continue with integration by parts.
Edit: Above has been corrected
So my question is: Have I done everything above correctly and how would I continue? Is there something I'm missing with this Uniform distribution because that integral doesn't seem to want to simplify.
Variance
Var(x) = E(x2) - [E(x)]2
And similarly for variance will follow.
Thanks!
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