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Is there an analogue to the higher moment Chebyshev's inequalities in the one sided case? The Chebyshev-Cantelli inequality only seem to work for the variance, whereas Chebyshevs' inequality can easily be produce for all exponents. Does anyone know of a one-sided inequality using the higher moments? |
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For convenience, let $X$ denote a continuous zero-mean random variable with density function $f(x)$, and consider $P\{X \geq a\}$ where $a > 0$. We have $$P\{X \geq a\} = \int_a^{\infty}f(x)\,\mathrm dx = \int_{-\infty}^{\infty}g(x)f(x)\,\mathrm dx = E[g(X)]$$ where $g(x) = \mathbf 1_{[a,\infty)}$. If $n$ is an even integer and $b$ any positive real number, then $$h(x) = \left(\frac{x+b}{a+b}\right)^n \geq g(x), -\infty < x < \infty,$$ and so $$E[h(X)] = \int_{-\infty}^{\infty} h(x)f(x)\,\mathrm dx \geq \int_{-\infty}^{\infty}g(x)f(x)\,\mathrm dx = E[g(X)].$$ Thus we have that for all positive real numbers $a$ and $b$, $$P\{X \geq a\} \leq E\left[\left(\frac{X+b}{a+b}\right)^n\right] = (a+b)^{-n}E[(X+b)^n]$$ where the rightmost expectation in the equation above is the $n$-th moment ($n$ even) of $X$ about $b$. When $n = 2$, the smallest upper bound on $P\{X \geq a\}$ is obtained when $b = \sigma^2/a$ giving the one-sided Chebyshev inequality (or Chebyshev-Cantelli inequality): $$P\{X \geq a\} \leq \frac{\sigma^2}{a^2 + \sigma^2}.$$ For larger values of $n$, minimization with respect to $b$ is messier. |
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