The Exponential Distribution

Please see a slightly expanded version here.

This is a derivation of the cumulative distribution function, characteristic function, moment generating function, first moment, expected value, second moment, and variance of the exponential distribution given its probability density function.

The probability density function of the exponential distribution is:

f_X(x) = \lambda e^{-\lambda x}

Thus the cumulative distribution function is:

F_X(x \leq a) = \displaystyle\int_{0}^{a} f_X(x) dx = \int_0^a \lambda e^{-\lambda x} dx = -e^{-\lambda x} \Big |_0^a = 1 - e^{-\lambda a}

And the characteristic function:

\phi_X(u) = E[e^{i u X}] = \displaystyle\int_{0}^{\infty} e^{i u x} f_X(x) dx = \int_0^{\infty} e^{i u x} \lambda e^{- \lambda x} dx = \lambda \int_0^{\infty} e^{-(\lambda - iu) x} dx = - \frac{\lambda}{\lambda-iu} e^{-(\lambda-iu)x} \Big |_0^{\infty} = \frac{\lambda}{\lambda - iu}

Now the moment generating function, which is obtained from the characteristic function:

M_X(t) = \phi_X(-it) = \displaystyle\frac{\lambda}{\lambda - (i(-it))} = \frac{\lambda}{\lambda - t}

Now the first moment:

M_X'(t) = \displaystyle\frac{d}{dt} \lambda(\lambda-t)^{-1} = -\lambda (\lambda -t)^{-2} \frac{d}{dt} (\lambda - t) = \frac{\lambda}{(\lambda - t)^2}


E[X] = M_X'(0) = \displaystyle\frac{1}{\lambda}

And the second moment:

M_X''(t) = \displaystyle\frac{d}{dt} M_X'(t) = \frac{d}{dt} \lambda (\lambda - t){-2} = -2 \lambda ( \lambda - t)^{-3} \frac{d}{dt} (\lambda - t) = \frac{2 \lambda}{(\lambda - t)^3}


E[X^2] = M_X''(0) = \displaystyle\frac{2}{\lambda^2}

And finally:

\displaystyle Var(X) = E[X^2] - E[X]^2 = \frac{2}{\lambda^2} - \Big(\frac{1}{\lambda}\Big)^2 = \frac{1}{\lambda^2}

Of course the expected value and the variance can be computed by evaluating:

\displaystyle \int_0^\infty x \lambda e^{-\lambda x} dx


\displaystyle \int_0^\infty x^2 \lambda e^{-\lambda x} dx


Next we’ll see how the exponential distribution is memoryless, that is:

\displaystyle P(X > t+s|X>t) = P(X > s)

First apply Bayes’ Rule:

\displaystyle P(X>t +s|X>t) = \displaystyle \frac{P(X>t|X>t+s)P(X>t+s)}{P(X>t)}
= \displaystyle \frac{1*P(X>t+s)}{P(X>t)}
= \displaystyle \frac{1-(1-e^{-\lambda (t+s)})}{1-(1-e^{-\lambda t})}
= \displaystyle \frac{e^{-\lambda (t+s)}}{e^{-\lambda t}}
= \displaystyle e^{-\lambda s}
= P(X>s)

Next, Let X_1, X_2 be exponential random variables with parameters \lambda_1 and \lambda_2 respectively, then:

\displaystyle P(X_1 < X_2) = \displaystyle \int_0^\infty P(X_1<X_2|X_1=x)P(X_1=x) dx
= \displaystyle \int_0^\infty P(X_1<X_2|X_1=x)\lambda_1 e^{-\lambda_1 x} dx
= \displaystyle \int_0^\infty P(x<X_2)\lambda_1 e^{-\lambda_1 x} dx
= \displaystyle \int_0^\infty P(X_2 \geq x)\lambda_1 e^{-\lambda_1 x} dx
= \displaystyle \int_0^\infty e^{-\lambda_2 x} \lambda_1 e^{-\lambda_1 x} dx
= \displaystyle \lambda_1 \int_0^\infty e^{-(\lambda_1 + \lambda_2)x} dx
= \displaystyle \lambda_1 \Big( -\frac{1}{\lambda_1+\lambda_2} e^{-(\lambda_1 + \lambda_2)x} \Big) \Big|_0^\infty
= \displaystyle \frac{\lambda_1}{\lambda_1+\lambda_2}


Ross, Sheldon M. Introduction to Probability Models, 9th edition. Academic Press. 2007.
Bremaud, Pierre. An Introduction to Probabilistic Modeling, 3rd printing. Springer. 1997.

  1. Leave a comment

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: