UA MATH564 概率不等式 QE练习题

• 2018年5月第三题

2018年5月第三题

Part a
If $x<t$, $1_{[t,+\infty )}(x)=0$,
$$1_{[t,+\infty )}(x)\le \frac{g(x)}{g(t)}\iff 0\le \frac{g(x)}{g(t)}$$

Notice $g$ is a non-decreasing function defined in $(0,+\infty )$, if $g(0^{+})$ is negative, then the inequality above may not hold for any fixed $t>0$ and arbitrary $x<t$. Hence, $g$ should be a positive function.

If $x\ge t$, $1_{[t,+\infty )}(x)=1$,
$$1_{[t,+\infty )}(x)\le \frac{g(x)}{g(t)}\iff 1\le \frac{g(x)}{g(t)}$$

Since $g$ is non-decreasing, $g(x)\ge g(t)$. Suppose $g$ is a positive function,
$$1_{[t,+\infty )}(x)\le \frac{g(x)}{g(t)}$$

Part b
Suppose $X$ is random variable defined on probability space $({\mathbb{R}}^{+},\mathcal{B}({\mathbb{R}}^{+}),P)$. By monotonicity of intergral
$${\int }_0^{\infty }{1}_{[t,+\infty )}(x)P(dx)\le {\int }_0^{\infty }\frac{g(x)}{g(t)}P(dx)$$

Notice
$$\begin{gathered} {\int }_0^{\infty }{1}_{[t,+\infty )}(x)P(dx)=P(X>t) \\ {\int }_0^{\infty }\frac{g(x)}{g(t)}P(dx)=\frac{Eg(X)}{g(t)} \end{gathered}$$

So
$$P(X>t)\le \frac{Eg(X)}{g(t)}$$

Part c
Let $g(x)=e^{sx}$ and then
$$Eg(Y)=E{e}^{sY}\le e^{\frac{{\sigma }^2{s}^2}{2}},\forall s\in \mathbb{R}$$

By inequality in Part b,
$$P(Y>t)\le \frac{Eg(Y)}{g(t)}\le \frac{e^{\frac{{\sigma }^2{s}^2}{2}}}{e^{st}}=e^{\frac{{\sigma }^2{s}^2}{2}-st}$$

When $s=\frac{t}{{\sigma }^2}$, $\frac{{\sigma }^2{s}^2}{2}-st$ reaches minimum $-\frac{t^2}{2{\sigma }^2}$. Inequality above holds for all $s\in \mathbb{R}$,
$$P(Y>t)\le \underset{s\in \mathbb{R}}{\inf }{e}^{\frac{{\sigma }^2{s}^2}{2}-st}=e^{-\frac{t^2}{2{\sigma }^2}}$$

总结

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