Exponential inverse cumulative distribution function

`X = expinv(P,mu)`

[X,XLO,XUP] = expinv(P,mu,pcov,alpha)

`X = expinv(P,mu)`

computes
the inverse of the exponential `cdf`

with parameters
specified by mean parameter `mu`

for the corresponding
probabilities in `P`

. `P`

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 and the values in `P`

must lie on the
interval [0 1].

`[X,XLO,XUP] = expinv(P,mu,pcov,alpha)`

produces confidence bounds for
`X`

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. `XLO`

and
`XUP`

are arrays of the same size as `X`

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 end points of that interval.

The inverse of the exponential cdf is

$$x={F}^{-1}(p|\mu )=-\mu \mathrm{ln}(1-p)$$

The result, *x*, is the value such that an
observation from an exponential distribution with parameter µ
will fall in the range [0 *x*] with
probability *p*.

Let the lifetime of light bulbs be exponentially distributed with µ = 700 hours. What is the median lifetime of a bulb?

expinv(0.50,700) ans = 485.2030

Suppose you buy a box of “700 hour” light bulbs. If 700 hours is the mean life of the bulbs, half of them will burn out in less than 500 hours.