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Let $X(t)$ be the standard Brownian motion, I need to find the distribution of $S=\int_{0}^T(X(t))^+dt$, where $(x)^+=\max\{0,x\}$.

I want to use the distribution to get a concentration bound for $S$. So even if we can't find the distribution, still, bounding the variance of $S$ helps a lot. For example, if I can bound the variance by, say $T^3$, then I can use Chebyshev's inequality to get a good concentration bound: \begin{align*} \Pr\{|S-E[S]|>k T^{3/2}\} < \frac{1}{k^2} \end{align*}

Any results on the same problem with $S=\int_{0}^T|X(t)|dt$ is also very much appreciated. Thank you!

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1 Answer 1

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The chances are that the distribution isn't pretty, but a decent concentration bound should be much easier. Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be convex and strictly increasing. Suppose we want to get a bound on $\mathbb{P}[g(S-\mathbb{E}S)\geq l]$, where $g:\mathbb{R}\to \mathbb{R}^+$ is convex. Use Markov's inequality to get

$$ \begin{align} \mathbb{P}[g(S-\mathbb{E}S)\geq l] &= \mathbb{P}[f(g(S-\mathbb{E}S))\geq f(l)]\\ &\leq \frac{\mathbb{E}[f(g(S-\mathbb{E}S))]}{f(l)}\\ &= \frac{\mathbb{E}[f(g(\int_0^T X_t^+-\mathbb{E}X_t^+dt))]}{f(l)}\\ &= \frac{\mathbb{E}[f(g(T\int_0^T X_t^+-\mathbb{E}X_t^+\frac{dt}{T}))]}{f(l)}\\ \end{align} $$

The composition of $f$ with $g$ is convex since $f$ is increasing, so, using Jensen's inequality

$$ \begin{align} \mathbb{P}[g(S-\mathbb{E}S)\geq l] &\leq \frac{\mathbb{E}[\int_0^T f(g(T(X_t^+-\mathbb{E}X_t^+)))\frac{dt}{T}]}{f(l)} \end{align} $$

Let $N$ be a standard normal random variable. Define $\varphi(m) = \mathbb{E}[f(g(m(N^+-\mathbb{E}N^+)))]$. Then

$$ \begin{align} \mathbb{P}[g(S-\mathbb{E}S)\geq l] &\leq \frac{\int_0^T \mathbb{E}[f(g(T(X_t^+-\mathbb{E}X_t^+)))]dt}{Tf(l)}\\ &\leq \frac{\int_0^T \varphi(T\sqrt{T})dt}{Tf(l)}\\ &= \frac{\varphi(T^{\frac{3}{2}})}{f(l)}\\ \end{align} $$

Example 1

Fix $g(x)=|x|$, corresponding to the problem of bounding $\mathbb{P}[|S-\mathbb{E}S|\geq l]$. Pick a function $f$ and see what bound you get. Take, for example,

$$ \begin{align} f(l) =e^{\frac{l^2}{4T^3}} \end{align} $$

Then, according to Mathematica,

$$ \begin{align} \varphi(T^{\frac{3}{2}}) &= \mathbb{E}\left[e^{\frac{(N^+-\mathbb{E}N^+)^2}{4}}\right]\\ &= 1.12 \\ &<\infty \\ \end{align} $$ So $$ \begin{align} \mathbb{P}[|S-\mathbb{E}S|\geq l] = O\left(e^{\frac{-l^2}{4T^3}}\right) \end{align} $$ If we let $l=kT^{\frac{3}{2}}$, then $$ \begin{align} \mathbb{P}[|S-\mathbb{E}S|\geq kT^{\frac{3}{2}}] = O\left(e^{\frac{-k^2}{4}}\right) \end{align} $$

Example 2

Let $g(x) = (-x)\mathbf{1}(x<0)$, corresponding to the problem of bounding $\mathbb{P}[\mathbb{E}(S)-S\geq l]$, for $l\geq 0$. Let $f(l)=\frac{l}{T^{\frac{3}{2}}}$. Then, by Mathematica,

$$ \begin{align} \varphi(m) &= T^{\frac{-3}{2}}\mathbb{E}[m(\mathbb{E}N^+-N^+)\mathbf{1}(N^+-\mathbb{E}N^+<0)]\\ &= T^{\frac{-3}{2}}m\left(\frac{1}{2 \sqrt{2 \pi }}+\frac{-1+e^{-\frac{1}{4 \pi }}+\frac{1}{2} \text{Erf}\left[\frac{1}{2 \sqrt{\pi }}\right]}{\sqrt{2 \pi }}\right)\\ &= 0.23m T^{\frac{-3}{2}} \end{align} $$

So

$$ \begin{align} \mathbb{P}[\mathbb{E}[S]-S\geq l]&\leq \frac{\int_0^T\varphi(t\sqrt{T})dt}{Tf(l)}\\ &= \frac{\varphi(1)\int_0^Tt\sqrt{T}dt}{Tf(l)}\\ &= \frac{0.12T^{\frac{3}{2}}}{l} \end{align} $$

So

$$ \begin{align} \mathbb{P}[\mathbb{E}[S]-S\geq kT^{\frac{3}{2}}]&\leq \frac{0.12}{k} \end{align} $$

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Thanks! It was a great bound, taught me a new technique and also works for the absolute value version! Awesome. –  afshi7n Dec 29 '12 at 18:54
    
Ben, actually, I noticed that the bound doesn't give much information in some cases: Since $E[S]=1/3 \sqrt{(2/\pi)} T^{3/2}$. To get a right-hand-side smaller than $1$, we should roughly have $k\geq 1$, but it makes the bound trivial since we get $\Pr[|S-1/3 \sqrt{2/\pi} T^{3/2}| \geq k T^{3/2}] = (1.12)e^{-1/4}$. It works great for when we want to bound $S$ from above (like Markov's inequality), but it doesn't give much information for when we want to bound $S$ from below. Maybe another choice of $f$ can make the argument work? Thank you very much for your time :-) –  afshi7n Jan 2 '13 at 2:30
    
@afshi7n Does this extension to the original idea give you a satisfactory bound? –  Ben Derrett Jan 2 '13 at 15:35
    
I'm amazed by seeing the flexibility of your solution! It gives a much better bound than what I could get before (using the second moment method and your first bound). One of the applications of these two bounds for me is computing a lower bound for $E[min\{E[S],S\}]$, using your second bound, I could prove that $0.19 E[S] \leq E[min\{E[S],S\}] $ (by integrating), this is already a great bound and maybe even close to a tight bound, but I still need to push it a little bit further and any comments or ideas on this are appreciated. You have been a great help already! Thanks a lot. –  afshi7n Jan 3 '13 at 21:13
    
@afshi7n Glad to help. We have 3 inequalities here: Markov's, Jensen's and the one in the integral (which is an equality in the second example). One thing you could do if you're in search of a better lower bound is to consider the second example and see how much probability "leaks" at each of the inequalities (via Monte Carlo simulation, perhaps). This might give you an idea for how to tighten things up. Let me know if you get a better bound! –  Ben Derrett Jan 3 '13 at 22:35

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