00:01
Hello students, here in the given problem, first to derive the maximum likelihood estimate of the lambda, here the sample observations that is x1, x2 up to xn from the poisson distribution here looking the likelihood distribution that is product of 1 to up to n e raised to minus lambda lambda raised to x upon x factorial.
00:32
Therefore, we can write it as e raised to minus n lambda lambda raised to summation of xi divided by product of i goes from 1 to n xi factorial.
00:45
Now, taking the log here minus n lambda plus summation of xi log of lambda minus summation of log of xi factorial.
00:59
Therefore, taking the derivative of with respect to lambda is equal to minus n plus summation of xi divided by lambda minus 0.
01:10
Therefore, here consider the derivative of lambda is equal to 0.
01:16
Therefore, the minus n plus summation of xi divided by lambda is equal to 0.
01:23
Therefore, we can write it as lambda hat is equal to summation of xi upon n that is x bar.
01:30
Therefore, we can say that the mle of lambda hat is summation of xi divided by n that is x bar.
01:41
Now, the second one to center limit theorem state that here for large enough samples size n the distribution of sample mean x bar will follows normal distribution with mean mu which is equal to population mean that is lambda and standard deviation of sigma which is equal to population standard deviation that is root lambda.
02:46
Therefore, here the deviation of the 95 percent confidence interval for lambda here by using the property of the normal distribution here x bar plus minus z into sigma upon root n.
03:10
Therefore, we can write it as x bar plus minus here 1 .96 into sigma upon root n because the 95 percent of confidence interval is 1 .96...