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I have the question:
Explain why a discrete uniform random variable, X, on [a,b] has the same variance as a discrete uniform random variable, W, on [1,b-a+1].
Use this information to establish that
http://thestudentroom.co.uk/latexren...93ea8adc71.png
...I don't even know where to start http://static.thestudentroom.co.uk/i...onal/frown.gif so any help would be great cheers
[a, b] --> [1, b - a + 1] suggests a horizontal translation of -a + 1: W = X - a + 1. And you should know that the shape of a distribution is unaffected by horizontal translation ....Quote:
Originally Posted by sirellwood
ah ok, so you are saying that Var(X)=(b-a)*(b-a)/12. and Var(W)=(b-a)*(b-a)/12
for the first one a=a, b=b; for the second one a=1, b=b-a+1; so...
Var(X1)=Var(W)?
$\displaystyle V(aX+b)=a^2V(X)$
The b, is a shift and that only changes the mean (center) not the variance (spread).
2b or not 2b? oh well it's 2am.
