Exponential CDFs, Survival Probabilities, and Quantiles

From the Density to Event Probabilities

For an exponential random variable, the main probability calculations come from integrating its density. The resulting CDF describes the probability of occurring by a given time; the survival function describes the probability of lasting beyond that time. Keeping those two events distinct is essential when translating a question into a calculation.

There are two equivalent parameterizations. With rate \(\lambda>0\), the rate is measured in inverse units of time. With scale \(\beta>0\), the scale has the same units as the random variable, and \(\beta=1/\lambda\). The exponential model has support \(x\geq 0\).

Definition: If \(X\) has an exponential distribution with rate \(\lambda>0\), its density is \(f(x)=\lambda e^{-\lambda x}\) for \(x\geq 0\), and \(f(x)=0\) for \(x<0\). Equivalently, with scale \(\beta=1/\lambda\), its density is \(f(x)=\frac{1}{\beta}e^{-x/\beta}\) for \(x\geq 0\). The CDF is \(F(x)=P(X\leq x)\); for \(x\geq0\), the survival function is \(S(x)=P(X>x)\).

Because an exponential distribution is continuous, \(P(X=x)=0\) for any fixed \(x\). Thus \(P(X>x)\) and \(P(X\geq x)\) are equal, as are \(P(X\leq x)\) and \(P(X<x)\). The distinction between the CDF and survival function is about which side of a threshold is being asked for, not whether the endpoint is included.

CDF and Survival Function

For \(x<0\), the random variable cannot take a value at or below \(x\), so its CDF is zero. For \(x\geq0\), integrate the density from the lower endpoint of the support to \(x\). The survival function can be calculated directly by integrating from \(x\) to infinity, or by taking the complement of the CDF.

Theorem: If \(X\) is exponential with rate \(\lambda>0\), then

F(x)= \begin{cases} 0, & x<0,\\ 1-e^{-\lambda x}, & x\geq0, \end{cases} \qquad S(x)=P(X>x)= \begin{cases} 1, & x<0,\\ e^{-\lambda x}, & x\geq0. \end{cases}

In scale notation, replace \(\lambda x\) by \(x/\beta\). In particular, for \(x\geq0\), \(F(x)=1-e^{-x/\beta}\) and \(S(x)=e^{-x/\beta}\).

Derivation. For \(x<0\), the density is zero on \((-\infty,x]\), giving \(F(x)=0\). For \(x\geq0\), direct integration gives

F(x)=\int_0^x \lambda e^{-\lambda t}\,dt =\left[-e^{-\lambda t}\right]_0^x =1-e^{-\lambda x}.

Likewise, for \(x\geq0\), integrating over the upper tail yields

S(x)=\int_x^\infty \lambda e^{-\lambda t}\,dt =\lim_{b\to\infty}\left(e^{-\lambda x}-e^{-\lambda b}\right) =e^{-\lambda x},

since \(\lambda>0\) implies \(e^{-\lambda b}\to0\) as \(b\to\infty\). The complement relation \(S(x)=1-F(x)\) gives the same result. Substituting \(\lambda=1/\beta\) gives the scale formulas. \(\square\)

These formulas also give interval probabilities. For \(0\leq a<b\),

P(a<X\leq b)=F(b)-F(a) =e^{-\lambda a}-e^{-\lambda b}.

For a lower-tail event, \(P(X\leq b)=F(b)\); for an upper-tail event, \(P(X>b)=S(b)\). If \(b<0\), the lower-tail probability is zero and the upper-tail probability is one.

Worked Example: An Upper Tail and an Interval Probability

Suppose a waiting time \(X\), measured in hours, is exponential with rate \(\lambda=0.2\) per hour. Find (a) the probability the wait exceeds 5 hours and (b) the probability it is more than 2 hours but no more than 8 hours.

(a) Exceeding 5 hours. This is an upper-tail event, so use the survival function:

P(X>5)=S(5)=e^{-0.2(5)}=e^{-1}\approx0.3679.

The probability is about \(0.3679\), or \(36.79\%\).

(b) Between 2 and 8 hours. Subtract the CDF at the lower endpoint from the CDF at the upper endpoint:

\begin{aligned} P(2<X\leq8) &=F(8)-F(2)\\ &=(1-e^{-0.2(8)})-(1-e^{-0.2(2)})\\ &=e^{-0.4}-e^{-1.6}\\ &\approx0.6703-0.2019\\ &\approx0.4684. \end{aligned}

As a check, this interval probability is positive because \(2<8\), and it is smaller than \(P(X>2)=e^{-0.4}\approx0.6703\), as it must be.

Quantiles and the Median

A quantile reverses the CDF calculation: instead of finding the probability by a given time, find the time corresponding to a specified cumulative probability. For a continuous, strictly increasing CDF on the positive support, there is a unique such time for every probability strictly between zero and one.

Definition: For \(0<p<1\), the \(p\)-quantile \(q_p\) of a distribution with a continuous, strictly increasing CDF is the unique value satisfying \(F(q_p)=p\). The median is the \(0.5\)-quantile.

For the exponential distribution, \(F(x)=1-e^{-\lambda x}\) is strictly increasing for \(x\geq0\), since its derivative is \(\lambda e^{-\lambda x}>0\). Solving \(F(q_p)=p\) gives the quantile directly.

Theorem: For an exponential random variable with rate \(\lambda>0\), the \(p\)-quantile and median are

q_p=-\frac{\ln(1-p)}{\lambda}, \qquad q_{0.5}=\frac{\ln 2}{\lambda}.

With scale \(\beta=1/\lambda\), these become \(q_p=-\beta\ln(1-p)\) and \(q_{0.5}=\beta\ln 2\).

Derivation. For \(0<p<1\), solve the CDF equation:

p=1-e^{-\lambda q_p} \quad\Longrightarrow\quad e^{-\lambda q_p}=1-p \quad\Longrightarrow\quad -\lambda q_p=\ln(1-p) \quad\Longrightarrow\quad q_p=-\frac{\ln(1-p)}{\lambda}.

Since \(1-p\) is strictly between zero and one, \(\ln(1-p)<0\), so \(q_p>0\), as required by the support. Substituting \(p=0.5\) gives \(q_{0.5}=-\ln(0.5)/\lambda=\ln(2)/\lambda\). Replacing \(1/\lambda\) by \(\beta\) gives the scale expressions. \(\square\)

Worked Example: Median and 90th Percentile in Scale Units

A device lifetime \(X\) is exponential with scale \(\beta=12\) minutes. Find its median and its 90th percentile, and check each by substituting it into the CDF.

Median. The scale-form median is \(\beta\ln2\), so

q_{0.5}=12\ln2\approx12(0.693147)=8.318\text{ minutes}.

Checking with the CDF, \(F(x)=1-e^{-x/12}\):

F(12\ln2)=1-e^{-(12\ln2)/12} =1-e^{-\ln2} =1-\frac12 =\frac12.

90th percentile. Set \(p=0.9\):

q_{0.9}=-12\ln(1-0.9) =-12\ln(0.1) =12\ln10 \approx27.631\text{ minutes}.

The check is \(F(12\ln10)=1-e^{-\ln10}=1-0.1=0.9\). Thus 90% of lifetimes are at or below about 27.63 minutes.

Solving for a Probability Threshold

Many questions specify a desired probability and ask for the time threshold that achieves it. These are quantile calculations when the condition is a lower-tail probability. For an upper-tail probability, solve the survival equation instead. The two approaches agree because \(S(t)=1-F(t)\).

Identify the event.
Decide whether the question asks for \(P(X\leq t)\) or \(P(X>t)\).

Choose the parameterization.
Use \(e^{-\lambda t}\) with a rate, or \(e^{-t/\beta}\) with a scale; check that the product or ratio in the exponent is dimensionless.

Isolate the exponential term.
Rearrange the probability equation, take natural logarithms, and solve for \(t\).

Check the answer.
Confirm that \(t\geq0\), the units are correct, and substitution returns the requested probability.

Worked Example: Find a Threshold in Rate and Scale Forms

First, suppose an event time \(X\) has rate \(\lambda=0.15\) per day. Find the time \(t\) by which the event has occurred with probability \(0.8\).

The event is \(X\leq t\), so solve \(F(t)=0.8\):

1-e^{-0.15t}=0.8 \quad\Longrightarrow\quad e^{-0.15t}=0.2 \quad\Longrightarrow\quad t=-\frac{\ln(0.2)}{0.15} =\frac{\ln5}{0.15} \approx10.730\text{ days}.

Checking, \(F(t)=1-e^{-0.15(\ln5/0.15)}=1-e^{-\ln5}=1-\frac15=0.8\).

Now suppose a lifetime is described using scale \(\beta=4\) hours. Find \(t\) such that the probability of lasting longer than \(t\) is \(0.1\). This is a survival-probability equation:

e^{-t/4}=0.1 \quad\Longrightarrow\quad -\frac{t}{4}=\ln(0.1) \quad\Longrightarrow\quad t=-4\ln(0.1)=4\ln10\approx9.210\text{ hours}.

Substitution gives \(S(t)=e^{-(4\ln10)/4}=e^{-\ln10}=0.1\). Equivalently, the rate is \(\lambda=1/4\) per hour, and solving \(e^{-\lambda t}=0.1\) gives the same threshold.

Common Mistakes and Why the Distinction Matters

The CDF and survival function are complementary ways to describe the distribution, but they answer different threshold questions. A statement such as “at least 10 hours” calls for an upper tail; “within 10 hours” calls for a lower tail. The exponential formulas make either calculation short, provided the event is translated correctly.

Using the wrong tail. \(P(X>t)\) is \(e^{-\lambda t}\), not \(1-e^{-\lambda t}\). The latter is \(P(X\leq t)\).
Confusing rate and scale. The exponent is \(-\lambda t\) for rate and \(-t/\beta\) for scale. Since \(\lambda\) has units of inverse time and \(\beta\) has units of time, both exponents are dimensionless.
Taking the logarithm before isolating the exponential. For example, \(1-e^{-\lambda t}=p\) first becomes \(e^{-\lambda t}=1-p\); then take logarithms.
Reporting an impossible threshold. For \(0<p<1\), the quantile is positive. A negative result often signals a sign error in solving the logarithm.

Key takeaway: Use \(F(t)=1-e^{-\lambda t}\) for the probability of occurring by \(t\), and \(S(t)=e^{-\lambda t}\) for the probability of exceeding \(t\). Their inverse gives \(q_p=-\ln(1-p)/\lambda\), or \(-\beta\ln(1-p)\) in scale form.

Check Your Understanding

For each question, identify the relevant CDF or survival calculation before evaluating or solving it.

If \(X\) has rate \(\lambda=0.5\) per hour, express \(P(X>3)\) using the survival function.
For an exponential variable with scale \(\beta=5\), derive the median and verify it by substituting into the CDF.
Suppose \(X\) has rate \(\lambda=0.1\) per day. Solve for \(t\) such that \(P(X\leq t)=0.95\).
For scale \(\beta=8\), solve for \(t\) such that \(P(X>t)=0.25\).
Explain why \(P(X\leq t)\) and \(P(X>t)\) sum to one, and state which one is represented by \(1-e^{-\lambda t}\) for \(t\geq0\).