Exponential cumulative distribution function


p = expcdf(x,mu)
[p,plo,pup] = expcdf(x,mu,pcov,alpha)
[p,plo,pup] = expcdf(___,'upper')


p = expcdf(x,mu) computes the exponential cdf at each of the values in x using the corresponding mean parameter mu. x and mu can be vectors, matrices, or multidimensional arrays that all have the same size. A scalar input is expanded to a constant array with the same dimensions as the other input. The parameters in mu must be positive.

[p,plo,pup] = expcdf(x,mu,pcov,alpha) produces confidence bounds for p when the input mean parameter mu is an estimate. pcov is the variance of the estimated mu. alpha specifies 100(1 - alpha)% confidence bounds. The default value of alpha is 0.05. plo and pup are arrays of the same size as p containing the lower and upper confidence bounds. The bounds are based on a normal approximation for the distribution of the log of the estimate of mu. If you estimate mu from a set of data, you can get a more accurate set of bounds by applying expfit to the data to get a confidence interval for mu, and then evaluating expinv at the lower and upper endpoints of that interval.

[p,plo,pup] = expcdf(___,'upper') returns the complement of the exponential cdf at each value in x, using an algorithm that more accurately computes the extreme upper tail probabilities. You can use the 'upper' argument with any of the prior syntaxes.

The exponential cdf is


The result, p, is the probability that a single observation from an exponential distribution will fall in the interval [0 x].


collapse all

The following code shows that the median of the exponential distribution is µ*log(2).

mu = 10:10:60; 
p = expcdf(log(2)*mu,mu)
p = 1×6

    0.5000    0.5000    0.5000    0.5000    0.5000    0.5000

What is the probability that an exponential random variable is less than or equal to the mean, µ?

mu = 1:6;
x = mu;
p = expcdf(x,mu)
p = 1×6

    0.6321    0.6321    0.6321    0.6321    0.6321    0.6321

Extended Capabilities

C/C++ Code Generation
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Introduced before R2006a