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world:useful-bounds


2017/10/07 23:39 ·

## Bounds on the tail of the normal distribution.

If $X_1$ follows a standard normal distribution, then $$\PP(X_1>x)=\int_x^\infty \frac{1}{\sqrt{2\pi}} e^{-t^2/2} dt \leq \int_x^\infty \frac{t}{x} \frac{1}{\sqrt{2\pi}} e^{-t^2/2} dt =\frac{e^{-x^2/2}}{x\sqrt{2\pi}}$$

## Bounds on the tail of a max distribution

Assume that $(X_i)$ are iid. Denote $M_n=\max_{i\in [1:n]} X_{i}$. $$\PP(M_n>x)\leq \sum_{i=1}^n \PP(X_i>x) \leq n \PP(X_1>x)$$ In the case where the distribution of $X_1$ is standard normal, then $$\label{eq:bound:max} \PP(M_n>x) \leq n \frac{e^{-x^2/2}}{x\sqrt{2\pi}}$$ The bound is not bad in $x$ but not very nice in $n$.

## Bounds on the moments of a max distribution

Let $(X_i)$ be iid standard gaussian random variables. Then, by Jensen's inequality, for all $\lambda \in (0,1/2)$, $$\rme^{\lambda \PE[M_n^2]} \leq \PE[\rme^{\lambda M_n^2}] \leq n \PE[\rme^{\lambda X^2}]=n \int \frac{\rme^{-\frac{1-2\lambda}{2}x^2}}{\sqrt{2\pi}} \rmd x=n \lr{1-2\lambda}^{-1/2}$$ Taking the $\log$ and dividing by $\lambda$, we get: $$\PE[M_n^2] \leq \frac{\log n-2^{-1}\log(1-2\lambda)}{\lambda}$$ Choosing $\lambda$ such that $\log n=-2^{-1}\log(1-2\lambda)$ yields for $n \geq 2$, $$\PE[M_n^2] \leq \frac{4\log n}{1-n^{-2}}= 4 \log n +\frac{4 \log n}{n^2-1} \leq 4\log n +1$$ With a similar argument, we can show that $$\PE[M_n] \leq \sqrt{2 \log n}$$ Finally, a Markov inequality yields for all $x>0$ $$\PP(M_n > x) \leq \frac{\sqrt{2 \log n}}{x}$$ which is better than the previous bound \eqref{eq:bound:max} wrt $n$ but dramatic wrt $x$…

## Another inequality which can be useful for max distribution

Let $Z$ be a non-negative random variable on a probability space $(\Omega,\mcf,\PP)$ and assume that there exists a constant $c> 1$ such that for all $t\geq 0$, $$\PP(Z>t)\leq c\rme^{-2nt^2}$$ Then, $$\PE[Z] \leq \sqrt{ \frac{\log c +1}{2n} }$$

$\bproof$ \begin{align*} \PE[Z^2]&=\int_0^\infty \PP(Z>\sqrt{t})\rmd t \leq \int_0^\infty 1\wedge \lr{c\rme^{-2nt} } \rmd t =\frac{\log c}{2n} + \int_{\frac{\log c}{2n}} ^\infty c\rme^{-2nt} \rmd t =\frac{\log c}{2n} +\frac{1}{2n} \end{align*} The proof is completed by noting that $\PE[Z] \leq \sqrt{\PE[Z^2]}$ $\eproof$

### Some comments on the approach

It may seem a bit convoluted to bound $\PE[Z]$ using a bound of $\PE[Z^2]$. I tried using a direct proof. \begin{align*} \PE[Z]&=\int_0^\infty \PP(Z>t)\rmd t \leq \int_0^\infty 1\wedge \lr{c\rme^{-2nt^2} } \rmd t =\sqrt{\frac{\log c}{2n}} + \int_{\sqrt{\frac{\log c}{2n}}} ^\infty c\rme^{-2nt^2} \rmd t \\ &\leq \sqrt{\frac{\log c}{2n}} +\int_{\sqrt{\frac{\log c}{2n}}} ^\infty \frac{t}{\sqrt{\frac{\log c}{2n}}} c\rme^{-2nt^2} \rmd t = \sqrt{\frac{\log c}{2n}} +\lrb{\frac{c\rme^{-2nt^2}}{-4n\sqrt{\frac{\log c}{2n}}} }_{\sqrt{\frac{\log c}{2n}}}^\infty \\ &=\frac{1}{\sqrt{2n}} \lr{\sqrt{\log c}+ \frac{1}{2 \sqrt{\log c}}} \end{align*} The bound is less sharp because on the second line, we only apply a rough bound on the survival function of a Gaussian distribution. And not surprisingly, the resulting bound is less sharp that the previous one because: $\sqrt{a+1} \leq \sqrt{a}+\frac{1}{2\sqrt{a}}$.

### Maximal Kolmogorov inequality

Let $(M_k)_{k\in\nset}$ be a square integrable $(\mcf_k)_{k\in\nset}$-martingale. Then, $$\PP\lr{\sup_{k=1}^n |M_k| \geq \alpha} \leq \frac{\PE[M_n^2]}{\alpha^2}$$

$\bproof$

We provide a complete proof here but actually, we can also apply Doob's inequality below to the non-negative submartingale $M_n^2$

Let $\sigma=\inf\set{k\geq 1}{|M_k| \geq \alpha}$ with the convention that $\inf \emptyset=\infty$. Then, $$\PP\lr{\sup_{k=1}^n |M_k| \geq \alpha} =\PP(\sigma \leq n)=\PP\lr{|M_{\sigma \wedge n}| \geq \alpha} \leq \frac{\PE[M^2_{\sigma \wedge n}]}{\alpha^2} \label{eq:kolm:one}$$ We first rewrite the rhs using that $(M_{\sigma \wedge k})_{k\in\nset}$ is also a $(\mcf_k)_{k\in\nset}$-martingale. To see this last property, write $M_{\sigma \wedge k}=\indiacc{k\leq \sigma} M_{k}+\indiacc{k> \sigma}M_\sigma$, which implies \begin{equation*} \PE\lrb{M_{\sigma \wedge k}|\mcf_{k-1}}=\indiacc{k\leq \sigma}\underbrace{\PE\lrb{M_{k}|\mcf_{k-1}}}_{M_{k-1}}+\indiacc{k> \sigma} M_\sigma=M_{\sigma \wedge (k-1)} \end{equation*} Now, the rhs of \eqref{eq:kolm:one} can be written using \begin{align*} {\PE[M^2_{\sigma \wedge n}]}&= {\PE[M_1^2]+\sum_{k=1}^{n-1} \PE\lrb{\lr{M_{\sigma \wedge (k+1)} - M_{\sigma \wedge k} }^2}}\\ &={\PE[M_1^2]+\sum_{k=1}^{n-1} \PE\lrb{\lr{M_{k+1} - M_{k} }^2 \indiacc{\sigma > k}}}\\ &\leq {\PE[M_1^2]+\sum_{k=1}^{n-1} \PE\lrb{\lr{M_{k+1} - M_{k} }^2 }}={\PE[M_n^2]} \end{align*} $\eproof$

### Doob's inequalities

• (i) Let $(X_k)_{k\in\nset}$ be a non-negative $(\mcf_k)_{k\in\nset}$-supermartingale. Then,

$$\PP\lr{\sup_{k=1}^\infty X_k \geq \epsilon} \leq \PE[X_0]/\epsilon$$

• (ii) Let $(X_k)_{k\in\nset}$ be a non-negative $(\mcf_k)_{k\in\nset}$-submartingale. Then,

$$\PP\lr{\max_{k=1}^n X_k \geq \epsilon} \leq \PE[X_n]/\epsilon$$

#### Proof

Define $\tau_\epsilon=\inf \set{k\in \nset}{X_k \geq \epsilon}$ with the convention that $\inf \emptyset =\infty$. Then, \begin{align} \label{eq:fond} \epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon}= \epsilon \PP\lr{\tau_\epsilon \leq n}&\leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}] \end{align}

• We prove (i). From \eqref{eq:fond}, we have

$$\epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon} \leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}] \leq \PE[X_{\tau_\epsilon \wedge n}]=\PE\lrb{\PE[X_{\tau_\epsilon \wedge n}|\mcf_0]}\leq \PE[X_{\tau_\epsilon \wedge 0}]=\PE[X_0]$$ where we used in the second inequality that $(X_n)$ is non-negative and in the third inequality that $(X_{\tau_\epsilon \wedge n})$ is a supermartingale. The proof then follows by letting $n$ goes to infinity.

• We now turn to the proof of (ii). Using \eqref{eq:fond},

\begin{align*} \epsilon \PP\lr{\max_{k=1}^n X_k \geq \epsilon}&\leq \PE[X_{\tau_\epsilon} \indiacc{\tau_\epsilon \leq n}]=\sum_{k=0}^n \PE[X_k \indiacc{\tau_\epsilon=k}]\\ &\leq \sum_{k=0}^n \PE[\PE[X_n|\mcf_k] \indiacc{\tau_\epsilon=k}]=\sum_{k=0}^n \PE[X_n \indiacc{\tau_\epsilon=k}]=\PE[X_n \indiacc{\tau_\epsilon\leq n}] \leq \PE[X_n] \end{align*} where we used in the last inequality that $(X_n)$ is non-negative. The proof is completed. $\eproof$