00:01
Hi guys, suppose that x1 and x2 are independent trend and variable having a common mean mu.
00:07
So let lambda x1 plus 1 minus lambda times x2 will be used as an estimate for mu for some appropriate value of lambda.
00:23
So it's known that the variance of x1 is sigma 1 squared.
00:31
And the variance of x2 with sigma 2 squared.
00:38
So we need to find the variance of this, which is lambda x1 plus 1 minus lambda times x2.
00:48
Okay, so variance of lambda x1 plus 1 minus lambda x2.
00:59
Okay, so this is variance of lambda x1.
01:05
Plus variance of 1 minus lambda x2 okay so this is equal to lambda variance of x1 plus 1 minus lambda square times variance of x okay so this is lambda square sigma 1 squared plus 1 minus lambda all squared sigma to square.
01:41
Okay, now we need to find the value of y yields the estimate having the lowest possible variance...