00:01
Hello students, according to the given question we have to find k1 and k2, the coincidence k1 and k2 values.
00:12
So here, y1 and y2 are independent, independent unbiased estimators of mu.
00:31
So expectation of y1 will be equal to mu and expectation of the one will be equal to mu and expectation of y2 will be equal to mu then expectation of k1 into y1 plus k2 into y2 will be equal to k1 expectation of y1 plus k2 into expectation of y2 so which is equal to k1 mu plus k2 into mu so which is equal to k1 plus k2 into mu thus as k1 into y1 plus k2 into y2 is unbiased so then k1 plus k2 into mu will be equal to mu so then k1 plus k2 value will be equal to 1.
01:31
Now variance of y1 will be equal to sigma 1 square and variance of y2 will be equal to sigma 2 square.
01:41
So as variance of y1 is twice the variance of y2.
01:55
That means variance of y1 will be equal to 2 sigma 2 square.
02:02
So as y 1 and y2 are independent, so k1 into y1 and k2 into y2 is also independent.
02:12
So, variants of k1, y1 plus k2, y2 will be equal to variance of k1 into y1 plus variance of k2 y2, so which is equal to k1 square into variance of y1 plus k2 square into variance of y2.
02:39
So as y1, y2 are independent, the covariance term will be 0.
02:44
So then which is equal to k1 square into here we have variance of y1 is 2 sigma 2 square.
02:55
So here 2 sigma 2 square plus k2 square into sigma 2 square which is equal to 2 k1 square plus k2 square into sigma 2 square.
03:11
So here sigma 2 square value is greater than 0.
03:16
So this is minimized when 2k1 square plus k2 square is minimized...