1 (ii) Through the numerical studies.

When the initial value is 1/4. And the stop condition is |f(xi)|<10^-6.

## [1] 0.250000000 0.130202916 0.066513792 0.033625543 0.016906993 0.008477317
## [7] 0.004244647

When the initial value is 1/10. And the stop condition is |f(xi)|<10^-6.

## [1] 0.100000000 0.050833194 0.025631922 0.012870710 0.006449160 0.003228046

When the initial value is 1/20. And the stop condition is |f(xi)|<10^-6.

## [1] 0.050000000 0.025208325 0.012657117 0.006341909 0.003174306

When the initial value is 1/100. And the stop condition is |f(xi)|<10^-6.

## [1] 0.010000000 0.005008333 0.002506257

When the initial value is -1/100. And the stop condition is |f(xi)|<10^-6.

## [1] -0.010000000 -0.004991667 -0.002493757

When the initial value is -1/20. And the stop condition is |f(xi)|<10^-6.

## [1] -0.050000000 -0.024791675 -0.012344619 -0.006159611 -0.003076644

When the initial value is -1/10. And the stop condition is |f(xi)|<10^-6.

## [1] -0.100000000 -0.049166806 -0.024381963 -0.012141442 -0.006058436
## [6] -0.003026160

When the initial value is -1/4. And the stop condition is |f(xi)|<10^-6.

## [1] -0.250000000 -0.119797084 -0.058702883 -0.029064289 -0.014461751
## [6] -0.007213447 -0.003602387

2 (i)

## The MLE of lambda is 0.3995092

For these two graphs, this is reasonable to assume that data comes from a Poisson distribution.

3(i)

This graph is ploted by the data mx,x.
From the graph we can see that there must exists the linear relatonship between X and m(X). Futhermore, the relationship must be twice times.

3(ii)

Use kernel method to estimate the relationship, we have to choose the bandwidth “bn” that can make our estimate more better first.
I use the Cross Validtion method to choose the bn. And the candidate bn is that the interval been separate to 1000 parts.

bn<-seq(0,1,1/1000)[-c(1:10)]

## The best bn is 0.029

Now, we get the bn. Use that “bn”" to do the kernel method.
I choose the Eparechnikor kernel to be my kernel function.
And set x0=1 which is the parameter of the Locally polynomial regression method.
Then I get the relationship between x and mx is that

## mx= 3.809468 + 6.246302 * (X-1) + 5.053587 * (X-1)^2
## The mean square error is -0.07110036