i´m lost in the use of pdfExp(x,a,m), cdfExp(x,a,m). Given that the exponential is iqual to:

What are the location and mean parameter?

For the cdfexp you said that it´s equal to:

qthelp://aptech.com.gauss.13.0/doc/Equation40.png

I don´t see any m there?

How can I have the same values than Matlab whith his expcdf and exppdf functions?

Thank you.

## 1 Answer

0

The `pdfExp` and `cdfExp` in GAUSS calculate the two parameter exponential function. This is documented more clearly and accurately in the latest version of GAUSS.

For easy comparison, I will show a few different parameterizations in GAUSS code. First, the single-parameter exponential function with a rate parameter called `lambda`. This is the one which you posted above.

val = lambda * exp(-lambda * x);

Second, the single-parameter exponential function with a scale parameter `beta`. This scale parameter is the reciprocal of the rate parameter.

val = (1/beta) * exp(-x/beta);

Finally, the two-parameter exponential function with a scale parameter `beta` and a threshold parameter (or location parameter in a sense) `theta`. This is the function in GAUSS.

val = (1/beta) * exp(-(x - theta)/beta));

To calculate the single-parameter exponential function in GAUSS, you will always set the second input equal to 0. If you are thinking of the distribution in terms of the rate parameter, `lambda` then you will pass in the final value as the `1/lambda`. Here are some examples:

//x = 1.2, lambda = 0.5 val_1 = 0.5 * exp(-1.2*0.5); //x = 1.2, beta = 2 val_2 = 1/2 * exp(-1.2/2); //x = 1.2, beta = 2, theta = 0 val_3 = 1/2 * exp(-(1.2 - 0)/2); //x = 1.2, beta = 2, theta = 0 val_4 = pdfExp(1.2, 0, 2);

For each of these the answer should be: 0.27440582.

## Your Answer

## 1 Answer

The `pdfExp` and `cdfExp` in GAUSS calculate the two parameter exponential function. This is documented more clearly and accurately in the latest version of GAUSS.

For easy comparison, I will show a few different parameterizations in GAUSS code. First, the single-parameter exponential function with a rate parameter called `lambda`. This is the one which you posted above.

val = lambda * exp(-lambda * x);

Second, the single-parameter exponential function with a scale parameter `beta`. This scale parameter is the reciprocal of the rate parameter.

val = (1/beta) * exp(-x/beta);

Finally, the two-parameter exponential function with a scale parameter `beta` and a threshold parameter (or location parameter in a sense) `theta`. This is the function in GAUSS.

val = (1/beta) * exp(-(x - theta)/beta));

To calculate the single-parameter exponential function in GAUSS, you will always set the second input equal to 0. If you are thinking of the distribution in terms of the rate parameter, `lambda` then you will pass in the final value as the `1/lambda`. Here are some examples:

//x = 1.2, lambda = 0.5 val_1 = 0.5 * exp(-1.2*0.5); //x = 1.2, beta = 2 val_2 = 1/2 * exp(-1.2/2); //x = 1.2, beta = 2, theta = 0 val_3 = 1/2 * exp(-(1.2 - 0)/2); //x = 1.2, beta = 2, theta = 0 val_4 = pdfExp(1.2, 0, 2);

For each of these the answer should be: 0.27440582.