Tail Bounds for Products and Ratios by Threshold Splitting

Why Split a Tail Event?

A product can be large because one factor is large, because another factor is large, or because several factors are moderately large. A ratio can be large because its numerator is large, because its denominator is close to zero, or both. If the joint distribution is unknown, these observations still give useful bounds: choose a threshold for each factor and cover the extreme event by a union of marginal events. The resulting bounds require no independence.

The method is flexible rather than automatic. A poor threshold can give a bound larger than one, or much larger than a better choice. The aim is to choose thresholds using the marginal information available for the factors.

Splitting a Product Tail

Theorem: Let $X_1,\ldots,X_m$ be nonnegative random variables on the same probability space, with any dependence structure, and let $t>0$. For any positive thresholds $b_1,\ldots,b_m$ satisfying $\prod_{i=1}^m b_i=t$, $$ P\!\left(\prod_{i=1}^m X_i>t\right) \leq \sum_{i=1}^m P(X_i>b_i). $$ The right-hand side may be replaced by its minimum with $1$.

To prove the event inclusion behind the theorem, suppose that $X_i\leq b_i$ for every $i$. Nonnegativity then gives $$ \prod_{i=1}^m X_i\leq \prod_{i=1}^m b_i=t. $$ Consequently, if the product is greater than $t$, at least one $X_i$ must be greater than $b_i$. Thus $$ \left\{\prod_{i=1}^m X_i>t\right\} \subseteq \bigcup_{i=1}^m \{X_i>b_i\}. $$ The union bound gives the stated probability inequality. Since every probability is at most $1$, taking the minimum with $1$ is also valid. At no point is a joint probability replaced by a product of probabilities.

For two factors, one can choose any $a>0$ and set the second threshold to $t/a$:

P(XY>t)\leq P(X>a)+P(Y>t/a), \qquad X,Y\geq 0.

This is a family of bounds, one for each $a$. When marginal tail bounds are known as functions of their thresholds, a useful general strategy is to minimize their sum over $a>0$. For $m$ factors, the corresponding optimization is over positive $b_i$ subject to $\prod_i b_i=t$.

Worked Example: Two Factors With Only First Moments

Suppose $X,Y\geq 0$, $E[X]=2$, and $E[Y]=3$. Bound $P(XY>48)$ without assuming independence.

Markov's inequality gives $P(X>a)\leq 2/a$ and $P(Y>48/a)\leq 3a/48$ for $a>0$. Threshold splitting therefore yields $$ P(XY>48)\leq \frac{2}{a}+\frac{a}{16}. $$ To minimize this expression, differentiate with respect to $a$. Its derivative is $-2/a^2+1/16$, which is zero when $a^2=32$, so take $a=4\sqrt{2}$. The two terms are then equal: $$ \frac{2}{4\sqrt{2}}+\frac{4\sqrt{2}}{16} =\frac{1}{2\sqrt{2}}+\frac{\sqrt{2}}{4} =\frac{1}{\sqrt{2}}. $$ Hence $P(XY>48)\leq 1/\sqrt{2}$, about $0.7071$. This is a valid dependence-free bound. It need not be the best possible bound given additional information about the joint distribution.

Products of Absolute Values

The same reasoning applies to real-valued factors after taking absolute values. For $X_1,\ldots,X_m$ real-valued, the event $\left|\prod_i X_i\right|>t$ is exactly the event $\prod_i |X_i|>t$. Thus, for any positive $b_i$ with product $t$,

P\!\left(\left|\prod_{i=1}^m X_i\right|>t\right) \leq \sum_{i=1}^m P(|X_i|>b_i).

This form is convenient when the factors can have either sign. For nonnegative factors, absolute values are unnecessary. In either case, the threshold product must match the target $t$; otherwise the event inclusion used in the proof need not hold.

Worked Example: Three Factors With Exponential Marginal Tails

Let $X_1,X_2,X_3$ be nonnegative random variables, not necessarily independent, and suppose each has the exponential distribution with rate $1$. Then $P(X_i>b)=e^{-b}$ for $b\geq 0$. Bound $P(X_1X_2X_3>64)$.

Choose equal thresholds $b_1=b_2=b_3=4$, whose product is $64$. Threshold splitting gives $$ P(X_1X_2X_3>64) \leq P(X_1>4)+P(X_2>4)+P(X_3>4) =3e^{-4}. $$ Numerically, $e^{-4}\approx 0.01832$, so $3e^{-4}\approx 0.05495$. The calculation uses only the three marginal exponential tails, not independence.

Equal thresholds are natural here because the marginal tail functions are identical and the target is $4^3$. They are not guaranteed to be optimal for other targets or for unequal marginal distributions.

Splitting a Ratio Tail

Theorem: Let $A$ and $B$ be real-valued random variables with $P(B=0)=0$, and let $t>0$. For any $a>0$, $$ P\!\left(\frac{|A|}{|B|}>t\right) \leq P(|A|>a)+P(|B|<a/t). $$ The variables may be dependent.

For a direct proof, suppose $|A|\leq a$ and $|B|\geq a/t$. Since $a/t>0$, these inequalities imply $|B|>0$, and $$ \frac{|A|}{|B|}\leq \frac{a}{a/t}=t. $$ Thus, if $|A|/|B|>t$, either $|A|>a$ or $|B|<a/t$. Taking probabilities and applying the union bound proves the result. The term $P(|B|<a/t)$ measures the denominator's lower tail near zero; an upper-tail bound for the denominator would not control this source of a large ratio.

If $A$ and $B$ are nonnegative and $B>0$ almost surely, this also bounds $P(A/B>t)$. If the ratio is defined only on the event $B\neq 0$, the assumption $P(B=0)=0$ makes its probability unambiguous without needing a separate convention for division by zero.

Worked Example: Balancing a Numerator Tail and a Small Denominator

Suppose $A$ has the exponential distribution with rate $1$, $B$ is uniform on $(0,1)$, and their dependence is unspecified. Bound $P(A/B>10)$.

For $0<a<10$, the ratio bound gives $$ P(A/B>10)\leq P(A>a)+P(B<a/10) =e^{-a}+\frac{a}{10}. $$ Here $P(B<a/10)=a/10$, because $a/10$ lies between $0$ and $1$. The derivative of the bound is $-e^{-a}+1/10$, which vanishes at $a=\log 10$. Since $0<\log 10<10$, this is an allowed threshold. Substitution gives $$ e^{-\log 10}+\frac{\log 10}{10} =\frac{1+\log 10}{10} \approx 0.33026. $$ Therefore $P(A/B>10)\leq (1+\log 10)/10$. This is an upper bound for any joint distribution with the stated marginals; it does not assert that the displayed value is attained.

Choosing Thresholds and Interpreting the Bound

Threshold choice is part of the method, not a technical afterthought. For a product, increasing one threshold decreases the corresponding upper-tail probability but forces the other thresholds to decrease if their product must remain fixed. For a ratio, increasing $a$ makes the numerator-tail term smaller while enlarging the denominator's small-value event. A useful threshold balances these competing effects.

If explicit marginal bounds are available, substitute them before optimizing. For example, if $X,Y\geq 0$, $E[X]=\mu_X<\infty$, and $E[Y]=\mu_Y<\infty$, Markov's inequality and product splitting give, for $a>0$, $$ P(XY>t)\leq \frac{\mu_X}{a}+\frac{\mu_Y a}{t}. $$ When both means are positive, the two terms are equal at $a=\sqrt{\mu_X t/\mu_Y}$. The resulting bound is $2\sqrt{\mu_X\mu_Y/t}$, which can be capped at $1$. This optimization is valid, but the resulting bound may still be uninformative when it exceeds $1$ before capping, or when it is close to $1$. Threshold splitting supplies a way to combine marginal bounds; it does not guarantee a sharp answer.

Common Pitfall: Multiplying Marginal Probabilities

A tempting but generally invalid step is to multiply the probabilities of the factor events. For example, $P(X>a,\,Y>b)\leq P(X>a)P(Y>b)$ is not valid without an appropriate independence assumption. Threshold splitting instead gives a union, whose probability is bounded by the sum of its marginal probabilities.

The difference can be seen with an explicit dependent example. Let $U$ be uniform on $(0,1)$, and set $X=Y=U$. Both $X$ and $Y$ are uniform on $(0,1)$, but they are perfectly dependent. Then $$ P(XY>1/4)=P(U^2>1/4)=P(U>1/2)=1/2. $$ Multiplying the two marginal probabilities would give $P(X>1/2)P(Y>1/2)=1/4$, which is too small to bound the true probability. Threshold splitting with $a=1/2$ gives $$ P(XY>1/4)\leq P(X>1/2)+P(Y>1/2)=1, $$ a valid, though less sharp, bound. In fact the union event here has probability $1/2$, matching the product-tail probability.

The central safeguard is simple: marginal tail probabilities can be added to bound a union regardless of dependence, but they cannot in general be multiplied to bound an intersection. Threshold splitting deliberately uses the first operation.

Key takeaway: For a product, choose factor thresholds whose product is the target. For a ratio, choose a numerator threshold and account for a denominator that falls below the corresponding scale. In both cases, combine marginal probabilities by a union bound, then tune the threshold to the available tail information.

Check Your Understanding

Use the event inclusions and marginal information in each question; do not assume independence unless it is stated.

For $X,Y\geq 0$, write a threshold-splitting bound for $P(XY>t)$ using an arbitrary threshold $a>0$.
Suppose $X_1,X_2,X_3\geq 0$. What condition must thresholds $b_1,b_2,b_3$ satisfy to bound $P(X_1X_2X_3>t)$ by their three marginal upper tails?
For $P(|A|/|B|>t)$, explain why the bound involves a lower-tail probability for $|B|$, rather than an upper-tail probability.
If $P(|A|>a)$ decreases as $a$ increases, what competing change occurs in the ratio bound's denominator term $P(|B|<a/t)$?
In the example $X=Y=U$ with $U$ uniform on $(0,1)$, why does multiplying $P(X>1/2)$ and $P(Y>1/2)$ fail to bound $P(XY>1/4)$?