00:01
So in this question, we're told that we have m coins, and they have probabilities.
00:13
So the probability that coin i is heads is i times a for i equals 1 to all the way up to m.
00:33
And we have that a is some fixed number between 0 and 1 over m.
00:40
So now let's say that we extract a coin from the jar, toss it, record the outcome and put it back in the jar.
00:49
So x .j is 1 or 0.
00:55
So this is if the jeth coin is heads, and this is if the jeth coin is tails.
01:10
Or the probability that xj is equal to 1 is going to be the sum.
01:18
Coin it is, so we have to sum over all m coins, we need to multiply by the probability that the coin is any one of those coins.
01:30
So the probability that xj is the, the coin is the ith coin, or the arth coin, it's 1 over m, because we're picking them at random.
01:41
And then the probability that the arth coin lands heads is r times a.
01:47
That's what we've got written above.
01:50
So we get a over m times the sum from r equals 1 to m of r.
01:56
Well, this is just an arithmetic series, which has sum m over 2, 1 plus m.
02:06
So the probability that the coin xj gives 1 is just a over 2, 1 plus m.
02:15
That means the expected value of xj is going to be 0 times the probability that xj equals 0 plus 1 times the probability that xj is 1, which is a over 2, 1 plus m.
02:34
And that means that the expected value of 2x1 over m plus 1 is going to be 2 over m plus 1 times the expected value of x1, which is 2 over m plus 1, a over 2, 1 plus m, which is a.
02:57
So the first estimator that we're given is an unbiased estimator of a.
03:04
For the rest of them, we're going to need to know what the expected value of kn is.
03:11
So the expected, so kn, the probability that kn, okay, so let's actually define kn first.
03:20
K -n is x1 plus x2 plus all the way up to x -n.
03:29
And the probability that k n is equal to j is going to be the probability that x i is equal to j is equal to 1, sorry, to the power of j, times the probability that x i is equal to zero to the power of n minus j, times n choose j, what we're saying is we have n choose j ways of choosing the j x's that we want to be equal to one.
04:19
And then for each of those configurations, we have to get j equal to one.
04:28
So that's the probability that they're equal to one to the power of j.
04:33
And then n minus j of them have to be equal to zero.
04:37
So that's why this factor is here.
04:40
So the expected value of kn is going to be the sum from j equals 1 up to n of n choose j, which is n factorial over j factorial n minus j factorial.
05:01
The probability that x i is equal to 1, which is a over 2, 1 plus m, and we raise that to the power j, and then we've got 1 minus a over 2, 1 plus m.
05:14
And then we've got 1 minus a over 2, 1 plus.
05:15
To the power of n minus j.
05:21
But actually, we need to multiply this by j because we're taking the expected value.
05:29
So we need to weight this sum with j.
05:32
But weighting it with j just means that this j factorial becomes j minus one factorial.
05:40
So now we can say that this is the sum from j equals zero to n.
05:49
So i'm replacing all the j minus ones with j's.
05:55
But this n minus j, becomes n minus 1 minus j.
06:02
And then this is going to be to the power of j plus 1.
06:10
And this is 1 minus a over 2, 1 plus m, n minus 1 minus j.
06:18
So now i can take out a factor of n and a factor of a over 2, 1 plus m.
06:28
And now i'm summing from j, sorry, this should be j equals 0 to n minus 1 here.
06:34
Because i'm just shifting the sum down by 1.
06:40
And this is n minus 1 factor.
06:42
Over j factorial, n minus 1 minus j factorial, a over 2, 1 plus m to the j, because i've taken a factor of this out, multiply by 1 minus a over 2, 1 plus m, to the n minus 1 minus j.
07:02
But if we look at this, this is n minus 1 choose j.
07:08
So i can get rid of this factor and write n minus 1 choose j...