The median of a probability distribution can be defined as the number m such that Upper P (Upper X less than or equals m )equals Upper P (Upper X greater than or equals m ). Find an expression for the median m of the exponential distribution.

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate).

Solution to the problem

For this case we can use the following Theorem:

"If X is a continuos random variable of the exponential distribution with parameter [tex]\beta[/tex] for some [tex]\beta \in R >0[/tex]"

Then the median of X is [tex]\beta ln (2)[/tex]

Proof

Let M the median for the random variable X.

From the definition for the exponential distribution we know the denisty function of X is given by:

[tex]f_X (x) = \frac{1}{\beta} e^{-\frac{x}{\beta}}[/tex]

Since we need the median we can put this equation:

[tex]P(X<M) = \frac{1}{\beta} \int_0^{M} e^{- \frac{x}{\beta}} dx = 1/2[/tex]

If we evaluate the integral we got this:

[tex]\frac{1}{\beta} \int_0^M e^{- \frac{x}{\beta}}dx =\frac{1}{\beta} [-\beta e^{-\frac{x}{\beta}}] \Big|_0^M[/tex]

And that's equal to:

[tex]1/2 = 1 -e^{- \frac{M}{\beta}}[/tex]

And if we solve for M we got:

[tex]1-e^{- \frac{M}{\beta}} = \frac{1}{2}[/tex]

[tex]e^{- \frac{M}{\beta}}=\frac{1}{2}[/tex]

If we apply natural log on both sides we got:

[tex]-\frac{M}{\beta}=ln(1/2)[/tex]

And then [tex]M=\beta ln(2)[/tex]