What the Survival Function Measures
A distribution can be described by asking how much probability lies below each threshold, as the cumulative distribution function does, or by asking how much lies above it. The latter view is especially direct when the event of interest is an exceedance: for example, whether a measurement is greater than a specified cutoff. The function that records these strict upper-tail probabilities is the survival function.
The distinction between “greater than” and “greater than or equal to” matters when the distribution assigns positive probability to an individual value. The survival function uses a strict inequality, so its value at a threshold does not include probability sitting exactly at that threshold.
This identity holds for every real \(x\), whether the distribution is discrete, continuous, or mixed. It also makes clear that the survival function is not a separate probability model: it is another way to represent the same distribution. Its values are between zero and one because they are probabilities.
Strict Exceedance and Point Masses
For any fixed \(x\), the event \(\{X\geq x\}\) includes the possibility \(X=x\), whereas \(\{X>x\}\) does not. The difference between these two probabilities is exactly the point mass at \(x\):
The CDF and survival function show the same point mass as a jump. Write \(F(x-)=\lim_{t\uparrow x}F(t)\) for the left-hand limit of the CDF. The probability at \(x\) is \(F(x)-F(x-)\). Correspondingly, define \(S(x-)=\lim_{t\uparrow x}S(t)\); the survival function’s downward jump is \(S(x-)-S(x)\).
Here is why the left limit of \(S\) has the stated interpretation. If \(t\) increases to \(x\), the events \(\{X>t\}\) decrease to \(\{X\geq x\}\). Continuity of probability for a decreasing sequence of events gives \(S(x-)=P(X\geq x)\). Subtracting \(S(x)=P(X>x)\) leaves \(P(X=x)\). This argument does not require a density or any other special form of the distribution.
Worked Example: A Discrete Distribution and Its Jumps
Suppose \(X\) takes the values \(-1\), \(0\), and \(2\) with probabilities \(0.2\), \(0.5\), and \(0.3\), respectively. Find its survival function and identify its jumps.
For \(x<-1\), all three possible values exceed \(x\), so \(S(x)=1\). For \(-1\leq x<0\), the value \(-1\) no longer exceeds \(x\), while \(0\) and \(2\) do, giving \(S(x)=0.5+0.3=0.8\). For \(0\leq x<2\), only \(2\) exceeds \(x\), giving \(S(x)=0.3\). At and above \(2\), no possible value exceeds \(x\), so \(S(x)=0\). Therefore,
At \(x=-1\), the left limit is \(1\) and the value is \(0.8\), so the downward jump is \(1-0.8=0.2=P(X=-1)\). At \(x=0\), it is \(0.8-0.3=0.5=P(X=0)\). At \(x=2\), it is \(0.3-0=0.3=P(X=2)\). The value at each point is the probability of being strictly greater than that point; this is why the probability at the point has already been excluded.
Monotonicity, Continuity, and Boundary Limits
The basic behavior of \(S\) follows from how exceedance events change with the threshold. Raising a threshold can only remove outcomes from the event. This observation proves monotonicity directly, while continuity of probability gives the appropriate one-sided continuity and limits.
- \(S\) is nonincreasing: if \(x_1\leq x_2\), then \(S(x_1)\geq S(x_2)\).
- \(S\) is right-continuous.
- \(\lim_{x\to-\infty}S(x)=1\) and \(\lim_{x\to+\infty}S(x)=0\).
- Its downward jump at any \(x\) equals \(P(X=x)\).
Proof. First suppose \(x_1\leq x_2\). If \(X>x_2\), then \(X>x_1\), so \(\{X>x_2\}\subseteq\{X>x_1\}\). Monotonicity of probability therefore gives \(S(x_2)\leq S(x_1)\).
For right continuity, take any sequence \(x_n\downarrow x\). The events \(\{X>x_n\}\) increase as \(n\) increases, and their union is \(\{X>x\}\). Indeed, if \(X>x\), then eventually \(x_n<X\); if \(X\leq x\), then \(X>x_n\) never holds. Continuity of probability for an increasing sequence of events gives
This sequential statement establishes right continuity on the real line. To obtain the limit at \(+\infty\), take \(x_n=n\). The events \(\{X>n\}\) decrease and their intersection is empty: a real-valued \(X\) cannot exceed every positive integer. Continuity of probability for decreasing events, whose first event has probability at most one, yields \(S(n)\to 0\). Since \(S\) is nonincreasing, this sequence limit implies \(\lim_{x\to+\infty}S(x)=0\).
For the other boundary, take \(x_n=-n\). The events \(\{X>-n\}\) increase and their union is the entire sample space, because every real value of \(X\) exceeds some negative integer. Thus \(S(-n)\to 1\) by continuity of probability for increasing events. Monotonicity then implies \(\lim_{x\to-\infty}S(x)=1\). Finally, the jump identity follows from \(S(x-)=P(X\geq x)\) and \(S(x)=P(X>x)\), so their difference is \(P(X=x)\). \(\square\)
The boundary limits express a simple fact: a finite real-valued observation eventually exceeds any threshold moved far enough to the left, and eventually fails to exceed a threshold moved far enough to the right. These limits do not say that the distribution has bounded support. A distribution can assign positive probability beyond every finite threshold and still have survival probabilities tending to zero as the threshold grows.
Worked Example: A Continuous Distribution
Let \(X\) have a uniform distribution on \((0,4)\). Its density is \(1/4\) on that interval and zero elsewhere. Find \(S(x)\), paying attention to the endpoints.
If \(x\leq 0\), then \(X>x\) with probability one, so \(S(x)=1\). If \(0<x<4\), the part of the interval \((0,4)\) above \(x\) has length \(4-x\), out of total length \(4\). Hence \(S(x)=(4-x)/4\). If \(x\geq 4\), then \(X>x\) has probability zero, so \(S(x)=0\). In full,
At \(x=4\), \(S(4)=0\) because the inequality is strict. The distribution has no point mass at \(4\), and the function reaches zero continuously there. It is also continuous at \(0\): both the value at \(0\) and the limit from the right are \(1\). The function decreases throughout the interval where the distribution has probability, in agreement with monotonicity.
Worked Example: A Mixed Distribution
Suppose \(X\) equals \(1\) with probability \(0.4\), and with probability \(0.6\) it has an exponential distribution with rate \(1\). The exponential component has survival probability \(e^{-x}\) for \(x\geq 0\). Find \(S(x)\) and check the jump at \(1\).
For \(x<0\), both the atom at \(1\) and every value from the exponential component exceed \(x\), so \(S(x)=1\). For \(0\leq x<1\), the atom at \(1\) exceeds \(x\), and the exponential component exceeds \(x\) with probability \(e^{-x}\). Weighting by their mixture probabilities gives \(S(x)=0.4+0.6e^{-x}\). At \(x=1\), the atom no longer counts because the condition is \(X>1\); only the exponential component contributes. For \(x\geq1\), therefore, \(S(x)=0.6e^{-x}\). Thus,
The left limit at \(1\) is \(0.4+0.6e^{-1}\), and the value at \(1\) is \(0.6e^{-1}\). Their difference is \(0.4\), exactly the probability assigned to \(X=1\). The continuous component contributes no additional point mass there.
Common Mistake: Switching the Inequality
A frequent error is to write \(S(x)=P(X\geq x)\) and then use \(S(x)=1-F(x)\). These two statements are not generally compatible. Since \(F(x)=P(X\leq x)\), its complement is \(P(X>x)\), not \(P(X\geq x)\). The correct relation is
For a continuous distribution with \(P(X=x)=0\), the strict and non-strict exceedance probabilities agree at that particular threshold. That agreement can make the error hard to notice. It need not hold for a discrete or mixed distribution, as the atom at \(1\) in the mixed example shows.
Check Your Understanding
Use the strict exceedance definition when answering.
- If \(P(X=3)=0.2\) and \(S(3-)=0.65\), what is \(S(3)\)?
- Why must \(S(x_2)\leq S(x_1)\) whenever \(x_1\leq x_2\)? State the event inclusion that justifies your answer.
- For \(X\) uniform on \((0,4)\), what is \(S(0)\), and why does it equal that value?
- Can a survival function have a downward jump at a point where \(P(X=x)=0\)? Explain using the jump identity.
- What are the limits of \(S(x)\) as \(x\) tends to negative and positive infinity for a real-valued random variable?