Load the sample data.

load carbig

The variable MPG has the miles per gallon for different models of cars.

Draw a histogram of MPG data.

 histogram(MPG)

The distribution is somewhat right skewed. A symmetric distribution, such as normal distribution, might not be a good fit.

Estimate the parameters of the Burr Type XII distribution for the MPG data.

phat = mle(MPG,'distribution','burr')

phat = 1×3

   34.6447    3.7898    3.5722

The maximum likelihood estimates for the scale parameter α is 34.6447. The estimates for the two shape parameters $$c$$ and $$k$$ of the Burr Type XII distribution are 3.7898 and 3.5722, respectively.

Generate sample data of size 1000 from a noncentral chi-square distribution with degrees of freedom 8 and noncentrality parameter 3.

rng default % for reproducibility
x = ncx2rnd(8,3,1000,1);

Estimate the parameters of the noncentral chi-square distribution from the sample data. To do this, custom define the noncentral chi-square pdf using the pdf input argument.

[phat,pci] = mle(x,'pdf',@(x,v,d)ncx2pdf(x,v,d),'start',[1,1])

phat = 1×2

    8.1052    2.6693

pci = 2×2

    7.1121    1.6025
    9.0983    3.7362

The estimate for the degrees of freedom is 8.1052 and the noncentrality parameter is 2.6693. The 95% confidence interval for the degrees of freedom is (7.1121,9.0983) and the noncentrality parameter is (1.6025,3.7362). The confidence intervals include the true parameter values of 8 and 3, respectively.

Load the sample data.

load('readmissiontimes.mat');

The data includes ReadmissionTime, which has readmission times for 100 patients. The column vector Censored has the censorship information for each patient, where 1 indicates a censored observation, and 0 indicates the exact readmission time is observed. This is simulated data.

Define a custom probability density and cumulative distribution function.

custpdf = @(data,lambda) lambda*exp(-lambda*data);
custcdf = @(data,lambda) 1-exp(-lambda*data);

Estimate the parameter, lambda, of the custom distribution for the censored sample data.

phat = mle(ReadmissionTime,'pdf',custpdf,'cdf',custcdf,'start',0.05,'Censoring',Censored)

phat = 0.1096

Load the sample data.

load('readmissiontimes.mat');

The data includes ReadmissionTime, which has readmission times for 100 patients. The column vector Censored has the censorship information for each patient, where 1 indicates a censored observation, and 0 indicates the exact readmission time is observed. This is simulated data.

Define a custom log probability density and survival function.

custlogpdf = @(data,lambda,k) log(k)-k*log(lambda)+(k-1)*log(data)-(data/lambda).^k;
custlogsf = @(data,lambda,k) -(data/lambda).^k;

Estimate the parameters, lambda and k, of the custom distribution for the censored sample data.

phat = mle(ReadmissionTime,'logpdf',custlogpdf,'logsf',custlogsf,...
    'start',[1,0.75],'Censoring',Censored)

phat = 1×2

    9.2090    1.4223

The scale and shape parameters of the custom-defined distribution are 9.2090 and 1.4223, respectively.

Load the sample data.

load('readmissiontimes.mat');

The data includes ReadmissionTime, which has readmission times for 100 patients. This is simulated data.

Define a negative log likelihood function.

custnloglf = @(lambda,data,cens,freq) - length(data)*log(lambda) + nansum(lambda*data);

Estimate the parameters of the defined distribution.

phat = mle(ReadmissionTime,'nloglf',custnloglf,'start',0.05)

phat = 0.1462

Generate 100 random observations from a binomial distribution with the number of trials, $$n$$ = 20, and the probability of success, $$p$$ = 0.75.

data = binornd(20,0.75,100,1);

Estimate the probability of success and 95% confidence limits using the simulated sample data.

[phat,pci] = mle(data,'distribution','binomial','alpha',.05,'ntrials',20)

phat = 0.7615

pci = 2×1

    0.7422
    0.7800

The estimate of probability of success is 0.7615 and the lower and upper limits of the 95% confidence interval are 0.7422 and 0.78. This interval covers the true value used to simulate the data.

Generate sample data of size 1000 from a noncentral chi-square distribution with degrees of freedom 10 and noncentrality parameter 5.

rng default % for reproducibility
x = ncx2rnd(10,5,1000,1);

Suppose the noncentrality parameter is fixed at the value 5. Estimate the degrees of freedom of the noncentral chi-square distribution from the sample data. To do this, custom define the noncentral chi-square pdf using the pdf input argument.

[phat,pci] = mle(x,'pdf',@(x,v,d)ncx2pdf(x,v,5),'start',1)

phat = 9.9307

pci = 2×1

    9.5626
   10.2989

The estimate for the noncentrality parameter is 9.9307, with a 95% confidence interval of 9.5626 and 10.2989. The confidence interval includes the true parameter value of 10.

Generate sample data of size 1000 from a Rician distribution with noncentrality parameter of 8 and scale parameter of 5. First create the Rician distribution.

r = makedist('Rician','s',8,'sigma',5);

Now, generate sample data from the distribution you created above.

rng default % For reproducibility
x = random(r,1000,1);

Suppose the scale parameter is known, and estimate the noncentrality parameter from sample data. To do this using mle, you must custom define the Rician probability density function.

[phat,pci] = mle(x,'pdf',@(x,s,sigma) pdf('rician',x,s,5),'start',10)

phat = 7.8953

pci = 2×1

    7.5405
    8.2501

The estimate for the noncentrality parameter is 7.8953, with a 95% confidence interval of 7.5404 and 8.2501. The confidence interval includes the true parameter value of 8.

Add a scale parameter to the chi-square distribution for adapting to the scale of data and fit it. First, generate sample data of size 1000 from a chi-square distribution with degrees of freedom 5, and scale it by the factor of 100.

rng default % For reproducibility
x = 100*chi2rnd(5,1000,1);

Estimate the degrees of freedom and the scaling factor. To do this, custom define the chi-square probability density function using the pdf input argument. The density function requires a $$1/s$$ factor for data scaled by $$s$$.

[phat,pci] = mle(x,'pdf',@(x,v,s)chi2pdf(x/s,v)/s,'start',[1,200])

phat = 1×2

    5.1079   99.1681

pci = 2×2

    4.6862   90.1215
    5.5297  108.2146

The estimate for the degrees of freedom is 5.1079 and the scale is 99.1681. The 95% confidence interval for the degrees of freedom is (4.6862,5.5279) and the scale parameter is (90.1215,108.2146). The confidence intervals include the true parameter values of 5 and 100, respectively.