= Tail probability
For a real <random variable> $X$, an upper-tail <probability> has the form $\mathbb P(X\geq t)$ and a lower-tail <probability> has the form $\mathbb P(X\leq t)$. Each is the <probability> of an <event> lying beyond a chosen threshold in one direction. A <tail probability> need not be small: the threshold determines how much of the <probability distribution> it includes. A <concentration inequality> bounds such <probabilities>, often for deviations from an <expected value>.
Endpoint conventions matter. Write $F(t)=\mathbb P(X\leq t)$ for the <cumulative distribution function> and $F(t^-)=\lim_{s\uparrow t}F(s)=\mathbb P(X<t)$. The <complement rule of probability> gives four distinct formulas:
$$
\begin{aligned}
\mathbb P(X\geq t)&=1-F(t^-),&\mathbb P(X>t)&=1-F(t),\\
\mathbb P(X\leq t)&=F(t),&\mathbb P(X<t)&=F(t^-).
\end{aligned}
$$
The strict upper-tail convention is the <survival function> $S(t)=\mathbb P(X>t)$. Its value differs from the inclusive upper <tail probability> by exactly the <probability mass> at the threshold:
$$
\mathbb P(X\geq t)-\mathbb P(X>t)=\mathbb P(X=t)=F(t)-F(t^-).
$$
A <singleton> with positive <probability mass> is an <atom of a measure>. Thus $1-F(t)$ excludes an <atom of a measure> at $t$, whereas $1-F(t^-)$ includes it. These endpoint identities follow directly from the <cumulative distribution function> properties in https://www.randomservices.org/random/dist/CDF.html[Kyle Siegrist, Distribution and Quantile Functions: endpoint probabilities and right tails].
As the threshold increases, both upper <tail probabilities> are <nonincreasing functions>. The strict <survival function> is a <right-continuous function>; the inclusive upper <tail probability> takes the upper endpoint value at a downward jump, equal to the left <limit> of the <survival function>. Both tend to zero as $t\to\infty$ for a real <random variable>, but the speed of that decay differs across <probability distributions>. The <cumulative distribution function> properties and their proof from continuity of <probability measures> are given in https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/9797310bed4c7f5b5d40d007783eec8d_MIT6_436JF18_lec04.pdf[MIT 6.436J, Lecture 4: random variables and cumulative distribution function properties].
For a fair six-sided die $D$, the <discrete uniform distribution> assigns <probability mass> $1/6$ to each of $1,\ldots,6$. At threshold four,
$$
\mathbb P(D\geq4)=\frac36=\frac12,\qquad \mathbb P(D>4)=\frac26=\frac13,\qquad F_D(4)=\frac46,\qquad F_D(4^-)=\frac36.
$$
Between four and five, both upper <tail probabilities> equal $1/3$. At the largest possible outcome the distinction is especially visible: $\mathbb P(D\geq6)=1/6$, but $\mathbb P(D>6)=0$. The diagram shows exact <probabilities> for this mathematical model; its dots specify which endpoints belong to each <step function>.
\Image[/tail-probabilities-die-endpoints.png]
{title=Exact fair-die probabilities: including or excluding a threshold changes the tail by one atom}
If a <probability distribution> has a <probability density function> $f$, then every <singleton> has zero <probability mass> and
$$
\mathbb P(X\geq t)=\mathbb P(X>t)=\int_t^\infty f(x)\,dx.
$$
For an <exponential distribution> of rate two, $f(x)=2e^{-2x}$ for $x\geq0$ and zero otherwise, so $S(t)=e^{-2t}$ for $t\geq0$ and $S(t)=1$ for $t<0$. Hence $\mathbb P(X\geq1)=e^{-2}\approx0.135335$. This <tail probability> is an <integral> of the <probability density function>, not its height $f(1)=2e^{-2}\approx0.270671$. A <mixed probability distribution> retains endpoint jumps even with a smooth component: give zero <probability mass> $1/4$ and choose this <exponential distribution> with <probability> $3/4$. Then $\mathbb P(X\geq0)=1$, $\mathbb P(X>0)=3/4$, and $S(t)=(3/4)e^{-2t}$ for $t\geq0$.
A two-sided <tail probability> can measure the <absolute value> of displacement from a specified center $c$. For $a>0$, the two extreme <events> are <disjoint events>, so
$$
\mathbb P(|X-c|\geq a)=\mathbb P(X\leq c-a)+\mathbb P(X\geq c+a)=F(c-a)+1-F((c+a)^-).
$$
For strict displacement $|X-c|>a$, replace the last expression by $F((c-a)^-)+1-F(c+a)$. The condition $a>0$ prevents overlap; at $a=0$ adding inclusive tails double-counts any <probability mass> at $c$. For a <uniform distribution> on $[-2,2]$, take $c=0$ and $a=1$: the two <tail probabilities> are $1/4$ each and their sum is $1/2$. Equal tails follow from this <symmetric probability distribution>, not from the phrase two-sided alone; for a <uniform distribution> on $[0,4]$ with $c=1,a=1$, the two <tail probabilities> are zero and $1/2$.
The <tail-sum formula for expectation> turns a whole curve of <tail probabilities> into an <expected value>. For any nonnegative <random variable> $Y$, without assuming a <probability density function> or a finite <expected value>,
$$
\mathbb E Y=\int_0^\infty\mathbb P(Y>t)\,dt=\int_0^\infty\mathbb P(Y\geq t)\,dt\quad\text{in }[0,\infty].
$$
For a proof outline, write $Y=\int_0^\infty\mathbf1_{\{Y>t\}}\,dt$ pointwise and exchange the <expected value> and <Lebesgue integral> using <Tonelli theorem>; $\mathbf1$ is an <indicator function>. The inclusive version has the same <Lebesgue integral>: positive <probability masses> occur at only a <countable set> of thresholds, which has <Lebesgue measure> zero. For a nonnegative <integer>-valued <random variable> $N$, the <tail-sum formula> instead gives $\mathbb E N=\sum_{k=1}^\infty\mathbb P(N\geq k)=\sum_{k=0}^\infty\mathbb P(N>k)$. The integer hypothesis is essential for these sums. The integral identity, its <indicator function> proof, and the integer version appear in https://www.randomservices.org/random/expect/Properties2.html[Kyle Siegrist, Additional Properties of Expected Value: integration and summation of right tails].
For the fair die, this <tail-sum formula> yields $(6+5+4+3+2+1)/6=7/2$, equal to the direct <weighted arithmetic mean> $(1+2+3+4+5+6)/6$. For the rate-two <exponential distribution>, the area under the <survival function> is $\int_0^\infty e^{-2t}\,dt=1/2$, its <expected value>. The first panel below shades an <integral> of the <probability density function> for a single threshold; the second shades an <integral> of the <survival function> across all nonnegative thresholds. These are analytical illustrations, with the displayed curves truncated at three; their exact total areas use an infinite integration range.
\Image[/tail-probabilities-density-and-expectation.png]
{title=Exponential model: one tail probability is density area, while expectation is total survival area}
Nonnegativity is essential to interpreting the upper <tail probability> area as $\mathbb E Y$: for a general real <random variable> that area equals the <expected value> of its <positive part>. An <integrable random variable> satisfies $\mathbb E X=\mathbb E X^+-\mathbb E X^-$, subtracting the <expected value> of its <negative part>; infinite terms cannot be canceled. Even when $Y$ is finite <almost surely>, its <expected value> can be infinite. For the shape-one <Pareto distribution> on $[1,\infty)$, $S(t)=1$ on $[0,1)$ and $S(t)=1/t$ for $t\geq1$, so the area is $1+\int_1^\infty dt/t=\infty$. The finite versus infinite <expected value> distinction is illustrated in https://www.randomservices.org/random/expect/Properties.html[Kyle Siegrist, Definitions and Basic Properties: exponential and Pareto expected values].
Why are <tail probabilities> useful? They answer threshold questions directly, and a <concentration inequality> can bound them without knowing the full <probability distribution>. For nonnegative $Y$ with finite <expected value>, <Markov inequality> gives $\mathbb P(Y\geq t)\leq\mathbb E Y/t$ for $t>0$: a large threshold relative to the <expected value> forces rarity. The <quantile function> reverses the question by choosing a threshold for a prescribed cumulative level. If $q=Q(1-\alpha)$ with $0<\alpha<1$, then $\mathbb P(X>q)\leq\alpha\leq\mathbb P(X\geq q)$; an <atom of a measure> may prevent equality. For the die, $Q(0.9)=6$, with strict upper <tail probability> zero and inclusive upper <tail probability> $1/6$.
A <tail probability> describes rarity under a specified <probability distribution>; it is not automatically a <conditional probability> after extra information. For $u\geq t$ with $\mathbb P(X\geq t)>0$,
$$
\mathbb P(X\geq u\mid X\geq t)=\frac{\mathbb P(X\geq u)}{\mathbb P(X\geq t)},
$$
because the threshold <events> are nested. Thus the die has $\mathbb P(D\geq5)=1/3$ but $\mathbb P(D\geq5\mid D\geq4)=2/3$. Conditioning changes the <sample space> and renormalizes its <probability measure>; the same extreme outcomes can become common in that restricted population.
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