Tightest Moment Bound on a Tail Probability
A positive random variable $X$ satisfies $E[X] = 1$ and $E[X^2] = 3$. Using only these two moments:
- Find the tightest possible upper bound on $P(X \geq 5)$. State which inequality you used.
- Construct (or describe) a distribution on $[0, \infty)$ that achieves the bound, proving it is tight.
Hints
- You have two moments, so you can do better than plain Markov. Think about which classical inequality uses both the mean and variance to bound a one-sided tail.
- Cantelli's inequality (the one-sided Chebyshev bound) states $P(X - \mu \geq t) \leq \sigma^2/(\sigma^2 + t^2)$. Compute the variance from the given moments and apply it with $t = 4$.
- To prove the bound is tight, construct a two-point distribution on $\{x_0, 5\}$ that matches both $E[X] = 1$ and $E[X^2] = 3$. Solve for the masses and the location $x_0$ -- you should find $x_0 = 1/2$ and $P(X = 5) = 1/9$.
Worked Solution
How to Think About It: You have two moments and you want to bound a tail probability. The first instinct is Markov's inequality, but you have a second moment -- you should be able to do better. The real question is: what is the sharpest bound you can get from both moments simultaneously? This is a classic moment problem, and the answer turns out to be Cantelli's inequality (the one-sided Chebyshev bound). To show it is tight, you construct a two-point distribution that saturates the bound.
Quick Estimate: Start with the easy bounds. Markov on $X$ gives $P(X \geq 5) \leq 1/5 = 0.20$. Markov on $X^2$ gives $P(X^2 \geq 25) \leq 3/25 = 0.12$. The variance is $\text{Var}(X) = 3 - 1 = 2$. Cantelli's inequality says $P(X - \mu \geq t) \leq \sigma^2/(\sigma^2 + t^2)$. Here $\mu = 1$, $\sigma^2 = 2$, $t = 4$, so $P(X \geq 5) \leq 2/(2 + 16) = 1/9 \approx 0.111$. This beats both Markov bounds, and as we will show, it is the best possible.
Approach: We solve the dual of the moment problem -- find the lowest-lying quadratic that dominates the indicator $\mathbf{1}_{\{x \geq 5\}}$ on $[0, \infty)$ -- and then verify sharpness by exhibiting an extremal distribution.
Formal Solution:
For any function $g(x) \geq \mathbf{1}_{\{x \geq 5\}}$ on $[0, \infty)$, we have $P(X \geq 5) \leq E[g(X)]$. Restrict to quadratics of the form $g(x) = c(x - x_0)^2$ with $c > 0$ and $x_0 \geq 0$. This ensures $g(x) \geq 0$ for all $x \geq 0$. Imposing $g(5) \geq 1$ gives $c = 1/(5 - x_0)^2$.
Then the bound becomes:
$$P(X \geq 5) \leq E[g(X)] = \frac{E[(X - x_0)^2]}{(5 - x_0)^2} = \frac{x_0^2 - 2x_0 + 3}{(5 - x_0)^2}$$
Minimize over $x_0$. Setting the derivative to zero:
$$\frac{d}{dx_0}\left[\frac{x_0^2 - 2x_0 + 3}{(5 - x_0)^2}\right] = \frac{(2x_0 - 2)(5 - x_0) + 2(x_0^2 - 2x_0 + 3)}{(5 - x_0)^3} = \frac{8x_0 - 4}{(5 - x_0)^3} = 0$$
This gives $x_0 = 1/2$, and the minimum value is:
$$\frac{(1/4) - 1 + 3}{(5 - 1/2)^2} = \frac{9/4}{81/4} = \frac{1}{9}$$
This is exactly Cantelli's inequality applied with $\mu = 1$, $\sigma^2 = 2$, $t = 4$.
Extremal distribution: Place mass at $x_0 = 1/2$ and at $x = 5$. Let $P(X = 5) = p$ and $P(X = 1/2) = 1 - p$. The moment constraints pin down $p$:
- $E[X] = \frac{1-p}{2} + 5p = \frac{1}{2} + \frac{9p}{2} = 1 \implies p = \frac{1}{9}$
- Check: $E[X^2] = \frac{1 - 1/9}{4} + 25 \cdot \frac{1}{9} = \frac{2}{9} + \frac{25}{9} = 3$ \checkmark
This distribution satisfies all constraints and achieves $P(X \geq 5) = 1/9$, so the bound is tight.
Answer: The tightest upper bound is $P(X \geq 5) \leq \dfrac{1}{9}$, obtained via Cantelli's (one-sided Chebyshev) inequality. It is achieved by the two-point distribution $P(X = 1/2) = 8/9$, $P(X = 5) = 1/9$.
Intuition
This problem is really about the duality between moment constraints and tail bounds. Every moment inequality (Markov, Chebyshev, Cantelli) works the same way: you find a function that dominates the indicator of the event you care about, then take expectations. The more moments you have, the richer the class of dominating functions, and the tighter the bound. Cantelli is what you get when you optimize over quadratics that are nonneg on the support -- it is the best you can do with just a mean and a variance for one-sided tails.
The extremal distribution is always illuminating. Here it is a two-point mass: most of the weight sits at $1/2$ (keeping the mean low and the second moment in check), and a thin spike at exactly $5$ carries all the tail probability. This is the adversarial distribution that makes the tail as heavy as possible while respecting the moment constraints. In practice, when you use moment bounds for risk management or pricing, you should think of these extremal distributions as the worst-case scenario your bound is protecting you against. If that worst case is unrealistic, your bound may be too conservative -- and that is useful information too.