Deriving the Exponential CDF and Survival Function

Why Derive Both Functions?

For a waiting time or lifetime, the cumulative distribution function answers how likely it is that the event has occurred by a specified time. The survival function answers how likely it is that the waiting time exceeds that time. Both functions come from the same exponential density, but the integral limits and the meaning of each result differ. Deriving them makes those distinctions explicit and helps prevent sign and support errors.

Assume throughout that \(T\) is a nonnegative waiting time with an exponential distribution of rate \(\lambda>0\). The rate is measured in inverse time units: if time is measured in months, for example, \(\lambda\) is per month. The density is zero for negative times and is \(\lambda e^{-\lambda t}\) for nonnegative times. This is a continuous model, so it assigns probability zero to any single time, including \(T=0\).

Deriving the Cumulative Distribution Function

By definition, \(F_T(t)\) is the probability that \(T\) lies at or below \(t\). Since \(T\) cannot be negative, \(F_T(t)=0\) whenever \(t<0\). For \(t\ge 0\), integrate the density over the possible values from 0 through \(t\):

The variable \(u\) is an integration variable; \(t\) is the time at which the CDF is evaluated. The antiderivative has a minus sign because the derivative of \(e^{-\lambda u}\) is \(-\lambda e^{-\lambda u}\). Consequently, the CDF for every real \(t\) is

At \(t=0\), the second expression gives \(1-e^0=0\), agreeing with the first branch’s limiting value. Thus the CDF is continuous at zero, as expected for a continuous distribution with no probability mass at that point. For \(t\ge0\), it starts at zero and increases toward one as \(t\) increases.

Deriving the Survival Function

For any real-valued random variable, the events \(T\le t\) and \(T>t\) are complementary. Therefore, \(S_T(t)=1-F_T(t)\). For \(t\ge0\), substituting the CDF gives

The same result follows directly by integrating the density above \(t\). For \(t\ge0\), the upper limit is infinity:

The term at infinity is zero because \(\lambda>0\), so \(e^{-\lambda u}\) tends to zero as \(u\) tends to infinity. For \(t<0\), all possible values of \(T\) exceed \(t\), so \(S_T(t)=1\). At \(t=0\), \(S_T(0)=P(T>0)=1\), because the model puts no probability mass at zero. Thus

For every \(t\), the two functions satisfy \(F_T(t)+S_T(t)=1\). The strict inequality in the survival definition causes no discrepancy here: \(P(T=t)=0\) for every fixed \(t\), so \(P(T>t)=P(T\ge t)\) under this continuous model.

Density, CDF, and Boundary Conditions

The CDF is a probability accumulated up to a threshold; the density is not itself a probability. For an absolutely continuous distribution, a density equals the derivative of the CDF almost everywhere. A density may be changed at a single point without changing any probability, so the general statement is not that a density must equal the derivative at every point.

For this exponential model, the displayed density \(\lambda e^{-\lambda t}\) equals the derivative of the displayed CDF at every \(t>0\). At the endpoint \(t=0\), treat the boundary separately: the CDF’s left and right derivatives do not agree there under the piecewise definition, and the value assigned to the density at that single point does not change the distribution. In particular, the density value \(f_T(t)\) has units of inverse time and should not be interpreted as \(P(T=t)\).

Using the Formulas Reliably

Before evaluating either function, identify the threshold’s sign and confirm that the time units match the rate. The exponent \(-\lambda t\) must be dimensionless. If time is in days, for instance, the rate used in the product must be per day. Unit consistency is part of the model calculation, not merely a reporting detail.

A useful check is to compare the result with the expected direction. For \(t\ge0\), \(F_T(t)=1-e^{-\lambda t}\) is between zero and one and increases with \(t\). In contrast, \(S_T(t)=e^{-\lambda t}\) is between zero and one and decreases with \(t\). Their sum must equal one at every threshold. These checks catch a reversed sign, an incorrect branch, or an accidental interchange of the functions.

Common Sign and Interpretation Traps

A frequent integration error is to write

This expression is negative for \(t>0\), so it cannot be a probability. The antiderivative is \(-e^{-\lambda u}\), not \(e^{-\lambda u}\). Applying the limits in order gives

Another error is to call \(e^{-\lambda t}\) the CDF. It is the survival function for nonnegative \(t\); the CDF is its complement. A final trap is to read a density value as the probability of an exact event time. For a continuous distribution, \(P(T=t)=0\), even though \(f_T(t)\) can be positive.