00:02
Ok, so we've got x1 to xn independent and identically distributed from the distribution, which is an exponential distribution with parameter theta.
00:19
And we want to find that this y is a sufficient and complete statistic.
00:25
So for a sufficient statistic, we need that the likelihood function for all the variables, variables and depending on theta factorizes as a function of theta that only depends on the data through this statistic y and then some other function which doesn't depend on theta but can depend on the data in any way we like and so here the likelihood function which since they're independent identically distributed is just the product of the individual pdfs which is theta to to the n e to the minus theta times the sum of the x's.
01:06
We can see that equals t of xi and s of theta y where t of xi is just one and s of theta and y is equal to e to the minus theta y times theta to the n.
01:25
So it is sufficient.
01:27
And is it complete? well for completeness we'd need that if the expectation of some function of y was 0 then that would imply that the probability that that function of y was 0 was equal to 1 i .e.
01:46
That if the expectation of the function equals 0 then the function equals 0 pretty much.
01:51
And so here we've got that if the expectation of g of y equals 0 that means that 0 is equal to the integral from 0 to infinity of g of y times theta to the n e to the minus theta y dy.
02:12
And we can see that this is the, if you take theta out the front, this is the laplace transform.
02:24
And so that equals zero if and only if g of y equals zero.
02:31
And therefore the probability that g of y equals zero does indeed equal 1.
02:39
And so it's complete 2.
02:47
For part b we're doing the expectation of n minus 1 over y, which is equal to n minus 1 times the expectation of 1 over y...