CDF of Exponential Distribution
The cdf of exponential distribution describes cumulative probabilities for a nonnegative waiting-time random variable. If events occur independently at a constant average rate, an exponential model is often used for the waiting time \(X\) until the next event.
Assume \(X \sim \text{Exponential}(\lambda)\) with rate parameter \(\lambda>0\). The support is \(x \ge 0\), and the mean waiting time is \(E[X]=\dfrac{1}{\lambda}\).
Step 1: Start from the Probability Density Function
The exponential probability density function (pdf) is:
Step 2: Integrate the PDF to Get the CDF
By definition, the cumulative distribution function is \(F(x)=P(X\le x)\). For \(x<0\), the probability is 0 because the exponential distribution has no mass on negative values. For \(x\ge 0\):
Therefore, the cdf of exponential distribution is:
Step 3: Use the CDF to Compute Common Probabilities
Once \(F(x)\) is known, many probabilities reduce to substitution and subtraction.
| Probability | Expression using the CDF | Simplified form |
|---|---|---|
| \(P(X\le x)\) for \(x\ge 0\) | \(F(x)\) | \(1-e^{-\lambda x}\) |
| \(P(X>x)\) for \(x\ge 0\) | \(1-F(x)\) | \(e^{-\lambda x}\) |
| \(P(a<X\le b)\) for \(0\le a<b\) | \(F(b)-F(a)\) | \(\left(1-e^{-\lambda b}\right)-\left(1-e^{-\lambda a}\right)=e^{-\lambda a}-e^{-\lambda b}\) |
| \(P(X\ge k)\) for integer-style wording, \(k\ge 0\) | \(P(X>k)\) (continuous) | \(e^{-\lambda k}\) |
For continuous random variables, \(P(X\ge x)=P(X>x)\) and \(P(X\le x)=P(X<x)\) because \(P(X=x)=0\).
Worked Example
Suppose the waiting time is exponential with rate \(\lambda=0.2\) per minute. Then \(E[X]=\dfrac{1}{0.2}=5\) minutes. Compute \(P(X\le 8)\) and \(P(3<X\le 8)\).
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Cumulative probability up to 8:
\[ P(X\le 8)=F(8)=1-e^{-0.2\cdot 8}=1-e^{-1.6}. \]
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Probability between 3 and 8:
\[ P(3<X\le 8)=F(8)-F(3)=\left(1-e^{-1.6}\right)-\left(1-e^{-0.2\cdot 3}\right) =e^{-0.6}-e^{-1.6}. \]
The exponential survival function is \(S(x)=P(X>x)=e^{-\lambda x}\). Many “at least” and “more than” waiting-time questions are most efficient using \(S(x)\) rather than summing areas directly.
Visualization: PDF Area Up to \(x\) Equals the CDF Value \(F(x)\)
Summary
The cdf of exponential distribution is obtained by integrating the pdf over \([0,x]\), giving \(F(x)=1-e^{-\lambda x}\) for \(x\ge 0\). Interval probabilities follow from differences of CDF values, and right-tail probabilities use the survival function \(e^{-\lambda x}\).