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Part-3, Q-3.6

S

shashankd

Member
Can anyone please explain why are we taking max|xi|, isn't |xi| enough ??
 
don't have the notes to hand, tho i do have exam papers, can you elaborate?
 
The likelihood function for the parameter Q based on a random sample of 'n'
observations from a population with a continuous uniform distribution on the range
(-Q / 2,Q / 2) is:
 
The density function for the data is
f(x,Q) = 1/2Q -Q <= x1, x2, x3…, xn <Q


So the likelihood is L(Q) = (1/2Q)^n

Since we can’t use calculus, we instead need to choose Q that maximises L(Q)

We need to choose the smallest value we can. But we need to satisfy the inequality above -Q <= x1, x2, x3…, xn <Q,

So the smallest Q we can choose is max |xi|
 
I think the density function should be 1/Q for -Q/2< xi < Q/2. So the likelihood function is (1/Q)^n.
 
I think the density function should be 1/Q for -Q/2< xi < Q/2. So the likelihood function is (1/Q)^n.

You're right, my solution assumed data on the range (-Q, Q). For the range (-Q/2, Q/2) adjust accordingly.
 
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