Rylan Schaeffer

Kernel Papers

Exponential Distribution


The exponential distribution is a single parameter \(\lambda\) continuous distribution over \((0, \infty)\). Its PDF is:

\[p(x)= \lambda e^{-\lambda x}\]

Its CDF is:

\[F(x) = \int_{0}^{x} \lambda e^{- \lambda t} dt = - e^{-lambda x} \lvert_{0}^{\infty} = 1 - e^{-\lambda x}\]

We can check that this is a valid CDF. As \(x\rightarrow \infty, F(x) \rightarrow 1\) and as \(x\rightarrow 0, F(x) \rightarrow 0\). Additionally, the CDF is monotonically increasing:

\[\partial_x F(x) = e^{-\lambda x} > 0\]



We say that a random variable \(X\) is memoryless if \(\forall s, t > 0\),

\[P(X\geq s + t \lvert X \geq s) = P(X \geq t)\]

Claim: Let \(X \sim Exp(\lambda)\). Then \(X\) is memoryless.


\[\begin{align*} P(X \geq t) &= 1 - F(t) = e^{-\lambda t}\\ P(X\geq s + t \lvert X \geq s) &= \frac{P(X\geq s + t, X \geq s)}{P(X \geq s)}\\ &= \frac{P(X\geq s + t)}{P(X \geq s)}\\ &= \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}}\\ &= e^{-\lambda t} \end{align*}\]

Claim: If \(X\) is a non-negative, continuous random variable that is memoryless, then \(X\sim Exp(\lambda)\) for some \(\lambda\). Intuitively, this means that the exponential distribution is the only memoryless distribution.


\(X\) being memoryless means that

\[\begin{align*} P(X \geq t) &= P(X\geq s + t \lvert X \geq s)\\ &= \frac{P(X\geq s + t, X \geq s)}{P(X \geq s)}\\ P(X \geq t) P(X \geq s) &= P(X\geq s + t) \end{align*}\]

This quantity, \(P(X \geq c)\) for some \(c\) is called the survival function \(S(\cdot) = 1 - F(\cdot)\). For our purposes, thinking in terms of the survival function is more clear. Here, we know that

\[S(s+t) = S(t)S(s)\]

Our strategy will be to find the survival function \(S(\cdot)\) and use it to show that \(X\) is exponential. Just by plugging in values, we can figure out a few things:

\[S(t + t) = S(2t) = S(t)S(t) = S(t)^2\] \[S(mt) = S(t)^m\] \[S(t/n) = S(t)^{1/n}\] \[S(\frac{m}{n}t) = S(t)^{m/n}\] \[\lim_{m/n \rightarrow c} S(\frac{m}{n}t)= S(\lim_{m/n \rightarrow c} \frac{m}{n}t) = S(t)^c\]

What function obeys \(S(ct) = S(t)^c\)? Setting \(t=1\) shows us that

\[S(x) = S(1)^x = e^{x \log S(1)}\]

Since \(S(1) = P(X \geq 1)\) is a probability, \(\log S(1) < 0\). Define \(- \lambda = \log S(1)\) with \(\lambda > 0\). Then we have

\[S(x) = e^{- \lambda x}\]

This is exactly the survival function of the exponential. Recall that the survival function was defined as \(S(x) = 1 - F(x)\) and the CDF of an exponential is \(F(x) = 1 - e^{-\lambda x}\). Consequently, we conclude that because \(X\) has the survival function of an exponential, it has the CDF of an exponential and is therefore an exponential random variable.