Say $X$ is a real random variable, and its expected value is $\mathbb{E}[X] = 0$. Denote the variance $\operatorname{Var}[X] = \sigma^2$.

Show that $P[X \geq \lambda] \leq \frac{\sigma^2}{\sigma^2 + \lambda^2}$ for any $\lambda > 0$.

This is an exercise in The Probabilistic Method (Alon & Spencer) in chapter 4, which focuses on the uses of the Chebyshev inequality - so I assume it is useful here but I'm stuck.


1 Answer 1


This is also Part (i) of Exercise 3.4 in Durrett, "Probability: Theory & Examples".

Chebyshev's inequality ((3.7) in the above) gives $i_A P(X \in A ) \le E \phi(X)$, where $\phi \ge 0$ and $i_A = \inf_{x \in A} \phi(x)$.

Let $\phi(x) = (x+ {\sigma^2 \over \lambda}) ^2$, with $A= \{ \omega | X (\omega) \ge \lambda \}$. Then $i_A = {(\sigma^2+\lambda^2)^2 \over \lambda^2}$, and $E \phi(X) = E [X^2 + 2X {\sigma^2 \over \lambda} + ({\sigma^2 \over \lambda})^2] = \sigma^2 +({\sigma^2 \over \lambda})^2= \sigma^2 { \sigma^2+\lambda^2 \over \lambda^2 }$, from which we get $P(X \in A ) \le {1 \over i_A} E \phi(X) = {\sigma^2 \over \sigma^2 + \lambda^2}$.


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