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Crossings in Stochastic Processes 📂Probability Theory

Crossings in Stochastic Processes

Definition

Let’s assume that we have a probability space (Ω,F,P)( \Omega , \mathcal{F} , P) and a submartingale {(Xn,Fn)}\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}. An upcrossing occurs when, over a closed interval [a,b][a,b], the value changes from being Xt1≤aX_{t_{1}} \le a to Xt2≥bX_{t_{2}} \ge b. The number of upcrossings observed up to time N∈NN \in \mathbb{N} is represented as follows: βN(a,b):=A number of upcrossing of {Xn} of interval [a,b] \beta_{N} (a,b): = \text{A number of upcrossing of } \left\{ X_{n} \right\} \text{ of interval } [a,b]

Basic Properties

  • [1]: χi\chi_{i} is a Fi−1\mathcal{F}_{i-1}-measurable function.
  • [2]: EβN(a,b)≤EXN++∣a∣b−a\displaystyle E \beta_{N} (a,b) \le {{ E X_{N}^{+} + |a| } \over { b-a }}
  • χi\chi_{i} being a Fi−1\mathcal{F}_{i-1}-measurable function means that for every Borel set B∈B(R)B \in \mathcal{B}(\mathbb{R}), χi−1(B)∈Fi−1\chi_{i}^{-1} (B) \in \mathcal{F}_{i-1} holds.

Explanation

20191014\_121834.png Simply put, an upcrossing refers to when XnX_{n} goes beyond the upper limit bb from the lower bound aa. The number of times this happens up to NN is represented as βN(a,b)\beta_{N} (a,b). In the above picture, it is βN(a,b)=3\beta_{N} (a,b) = 3.

  • [1]: Before explaining what χi\chi_{i} is, I want to introduce a few notations frequently used for upcrossings. If you don’t like to read these explanations, it’s okay to just look at the pictures and understand intuitively. τ1:=min⁡n{n≤N:Xn≤a}τ2:=min⁡n{τ1<n≤N:Xn≥b}τ3:=min⁡n{τ2<n≤N:Xn≤a}τ4:=min⁡n{τ2<n≤N:Xn≥b}⋮ \tau_{1}:= \min_{n} \left\{ \qquad n \le N: X_{n} \le a \right\} \\ \tau_{2}:= \min_{n} \left\{ \tau_{1} < n \le N: X_{n} \ge b \right\} \\ \tau_{3}:= \min_{n} \left\{ \tau_{2} < n \le N: X_{n} \le a \right\} \\ \tau_{4}:= \min_{n} \left\{ \tau_{2} < n \le N: X_{n} \ge b \right\} \\ \vdots Defined as above, τk\tau_{k} represents a stopping time when the random variable XnX_{n} exits the interval [a,b][a,b]. According to the definition, for odd kk, it exits below aa, and for even kk, it exits above bb. Thus, typically, for a natural number mm, the moment it exits below is τ2m−1\tau_{2m-1}, and the moment it exits above is τ2m\tau_{2m}. Using this expression, mm can naturally represent the mmth upcrossing. 20191014\_122257.png

JmJ_{m} represents the set of indices where the mmth upcrossing is occurring. If written mathematically, it’s as follows: Jm:={k∈N:τ2m−1+1≤k≤τ2m} J_{m}:= \left\{ k \in \mathbb{N}: \tau_{2m-1} + 1 \le k \le \tau_{2m} \right\} Regarding this, χi\chi_{i} only takes the value of 11 when an upcrossing is happening, and otherwise, it’s 00. The intent behind using such a function is to isolate only the parts where upcrossings occur and eliminate the rest by multiplying them with 00. Mathematically, it’s defined as the following indicator function: χi=1⋃Jm={0,i∈J1∪⋯∪Jm1,otherwise \begin{align*} \chi_{i} =& \mathbb{1}_{ \bigcup J_{m} } \\ =& \begin{cases} 0 &, i \in J_{1} \cup \cdots \cup J_{m} \\ 1 &, \text{otherwise} \end{cases} \end{align*} Let’s visually check them in the following picture: 20191014\_122836.png

  • [2]: While we can’t precisely calculate EβN(a,b)E \beta_{N} (a,b), being able to compute its upper limit is quite good. A notable point here is that we don’t need to observe all cases, but just calculate the last occurrence EXN+E X_{N}^{+}.

Proof

[1]

Part 1. (τ2m−1<i≤τ2m)=(τ2m−1<i)∩(i≤τ2m)(\tau_{2m-1} < i \le \tau_{2m}) = (\tau_{2m-1} < i ) \cap ( i \le \tau_{2m})

χi\chi_{i} can be represented as follows, according to the definition: χi=1⋃Jm=∑m=1βN(a,b)1Jm=∑m=1βN(a,b)1(τ2m−1<i≤τ2m) \begin{align*} \chi_{i} =& \mathbb{1}_{\bigcup J_{m}} \\ =& \sum_{m=1}^{\beta_{N} (a,b) } \mathbb{1}_{J_{m}} \\ =& \sum_{m=1}^{\beta_{N} (a,b) } \mathbb{1}_{(\tau_{2m-1} < i \le \tau_{2m})} \end{align*} So, we need to check if it’s (τ2m−1<i≤τ2m)∈Fi−1(\tau_{2m-1} < i \le \tau_{2m}) \in \mathcal{F}_{i-1}. If we dissect it as an intersection: (τ2m−1<i≤τ2m)=(τ2m−1<i)∩(i≤τ2m) (\tau_{2m-1} < i \le \tau_{2m}) = (\tau_{2m-1} < i ) \cap ( i \le \tau_{2m})

Part 2. (i≤τ2m)∈Fi−1(i \le \tau_{2m} ) \in \mathcal{F}_{i-1}

From the definition of τk\tau_{k}, k=2mk = 2m, i.e., the case of even numbers, represents the moment when XnX_{n} exceeds bb. However, for an upcrossing to occur, XnX_{n} must go from below aa to above bb, which means returning to a point below aa takes at least one step. Thus, if χi−1=1\chi_{i-1} = 1 and Xi−1≥bX_{i-1} \ge b, the next step inevitably must be xi=0x_{i} = 0. This is almost as if we have determined χi\chi_{i}, even though we are only observing up to i−1i-1 with just Fi−1\mathcal{F}_{i-1} worth of information. Therefore, it’s (i≤τ2m)∈Fi−1(i \le \tau_{2m} ) \in \mathcal{F}_{i-1}.

Part 3. (τ2m−1<i)∈Fi−1(\tau_{2m-1} < i ) \in \mathcal{F}_{i-1}

According to the definition of the sigma field Fi−1\mathcal{F}_{i-1}: (i≤τ2m)∈Fi−1  ⟹  (i≤τ2m)c∈Fi−1  ⟹  (τ2m<i)∈Fi−1  ⟹  (τ2m−1<i)∈Fi−1 \begin{align*} & (i \le \tau_{2m} ) \in \mathcal{F}_{i-1} \\ \implies& (i \le \tau_{2m} )^{c} \in \mathcal{F}_{i-1} \\ \implies& (\tau_{2m} < i ) \in \mathcal{F}_{i-1} \\ \implies& (\tau_{2m-1} < i ) \in \mathcal{F}_{i-1} \end{align*}

Part 4. (τ2m−1<i≤τ2m)∈Fi−1(\tau_{2m-1} < i \le \tau_{2m}) \in \mathcal{F}_{i-1}

According to the definition of the sigma field Fi−1\mathcal{F}_{i-1}: (τ2m−1<i)∈Fi−1∧(i≤τ2m)∈Fi−1  ⟹  (τ2m−1<i)∩(i≤τ2m)∈Fi−1  ⟹  (τ2m−1<i≤τ2m)∈Fi−1 \begin{align*} & (\tau_{2m-1} < i ) \in \mathcal{F}_{i-1} \land (i \le \tau_{2m} ) \in \mathcal{F}_{i-1} \\ \implies& (\tau_{2m-1} < i ) \cap ( i \le \tau_{2m}) \in \mathcal{F}_{i-1} \\ \implies& (\tau_{2m-1} < i \le \tau_{2m}) \in \mathcal{F}_{i-1} \end{align*}

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[2]

βN(a,b)=A number of upcrossing of {Xn} of interval [a,b]=A number of upcrossing of {Xn−a} of interval [0,b−a]=A number of upcrossing of {(Xn−a)+} of interval [0,b−a] \begin{align*} \beta_{N} (a,b) =& \text{A number of upcrossing of } \left\{ X_{n} \right\} \text{ of interval } [a,b] \\ =& \text{A number of upcrossing of } \left\{ X_{n} - a \right\} \text{ of interval } [0,b-a] \\ =& \text{A number of upcrossing of } \left\{ ( X_{n} - a )^{+} \right\} \text{ of interval } [0,b-a] \end{align*} Therefore, by proving the following inequality for Yn:=(Xn−a)+Y_{n}:= ( X_{n} - a )^{+}, we can assert that EβN(a,b)≤EXN++∣a∣b−a\displaystyle E \beta_{N} (a,b) \le {{ E X_{N}^{+} + |a| } \over { b-a }} holds without loss of generality: EβN(0,b)≤EYNb E \beta_{N} (0,b) \le {{ E Y_{N}} \over { b }}

Part 1. E(Yn+1∣Fn)≥YnE \left( Y_{n+1} | \mathcal{F}_{n} \right) \ge Y_{n}

20191014\_123800.png

Looking at the above figure, it’s easy to see that f(x)=(x−a)+f(x) = (x - a)^{+} is a convex function and is non-decreasing. Thus, according to the conditional Jensen’s inequality: E(Yn+1∣Fn)=E((Xn+1−a)+∣Fn)≥(E(Xn+1−a∣Fn))+≥(Xn−a)+=Yn \begin{align*} E \left( Y_{n+1} | \mathcal{F}_{n} \right) =& E \left( ( X_{n+1} - a )^{+} | \mathcal{F}_{n} \right) \\ \ge& \left( E \left( X_{n+1} - a | \mathcal{F}_{n} \right) \right)^{+} \\ \ge& \left( X_{n} - a \right)^{+} \\ =& Y_{n} \end{align*}

Part 2. bEβN(0,b)≤EYN\displaystyle b E \beta_{N} (0,b) \le E Y_{N}

20191014\_124310.png Regardless of the length bb of the interval [0,b][0,b], from the definition of τk\tau_{k}, τ2m\tau_{2m} means the point when it exceeds bb above, so it’s Yτ2m≥bY_{\tau_{2m}} \ge b, and τ2m−1\tau_{2m-1} means the moment it falls below YkY_{k}, so it’s Yτ2m≤0Y_{\tau_{2m}} \le 0. Thus, the increase in Yk Y_{k} during the mmth upcrossing is Y2m−Y2m−1≥bY_{2m} - Y_{2m-1} \ge b, and this always holds whenever there’s an upcrossing, for βN[0,b]\beta_{N} [0,b] times. Therefore, bβN[0,b]b \beta_{N} [0,b] must be equal to or less than each of these increases. If we represent it as a formula, it’s as follows: bβN(0,b)≤∑m=1βN(0,b)(Yτ2m−Yτ2m−1)=∑m=1βN(0,b)[(Yτ2m−Yτ2m−1)+(Yτ2m−1−Yτ2m−2)+⋯+(Yτ2m−1+1−Yτ2m−1)]=∑m=1βN(0,b)∑i∈Jm(Yi−Yi−1) \begin{align*} b \beta_{N} (0,b) \le & \sum_{m=1}^{\beta_{N} (0,b)} \left( Y_{\tau_{2m}} - Y_{\tau_{2m-1}} \right) \\ =& \sum_{m=1}^{\beta_{N} (0,b)} \left[ \left( Y_{\tau_{2m}} - Y_{\tau_{2m}-1} \right) + \left( Y_{\tau_{2m}-1} - Y_{\tau_{2m}-2} \right) + \cdots + \left( Y_{\tau_{2m-1}+1} - Y_{\tau_{2m-1}} \right) \right] \\ =& \sum_{m=1}^{\beta_{N} (0,b)} \sum_{i \in J_{m}} \left( Y_{i} - Y_{i-1} \right) \end{align*} Because we don’t like counting indices as βN(0,b)\beta_{N} (0,b), instead of looking at the whole i=1,⋯ ,Ni=1 , \cdots , N, we use χi\chi_{i}, which multiplies by 11 while upcrossing is occurring and by 00 when it’s not happening. Then, the above formula simplifies as follows: bβN(0,b)≤∑m=1βN(0,b)∑i∈Jm(Yi−Yi−1)=∑i=1N(Yi−Yi−1)χi b \beta_{N} (0,b) \le \sum_{m=1}^{\beta_{N} (0,b)} \sum_{i \in J_{m}} \left( Y_{i} - Y_{i-1} \right) = \sum_{i=1}^{N} ( Y_{i} - Y_{i-1}) \chi_{i}

Part 3.

Properties of Conditional Expectation:

  • [3]: If XX is F\mathcal{F}-measurable, then E(X∣F)=X a.s.E(X|\mathcal{F}) =X \text{ a.s.}
  • [11]: For all sigma fields G\mathcal{G}, hence E[E(X∣G)]=E(X)E \left[ E ( X | \mathcal{G} ) \right] = E(X)

Taking the expectation, according to [11], for the conditional expectation E[⋅∣Fi−1]E \left[ \cdot | \mathcal{F}_{i-1} \right]: bEβN(0,b)≤∑i=1NE(Yi−Yi−1)χi=∑i=1NE[E[(Yi−Yi−1)χi∣Fi−1]] \begin{align*} b E \beta_{N} (0,b) \le & \sum_{i=1}^{N} E ( Y_{i} - Y_{i-1}) \chi_{i} \\ &\color{red}{=}& \sum_{i=1}^{N} E \left[ E \left[ ( Y_{i} - Y_{i-1}) \chi_{i} | \mathcal{F}_{i-1} \right] \right] \end{align*}

Smoothing Property of Conditional Expectation: If XX is G\mathcal{G}-measurable, then E(XY∣G)=XE(Y∣G) a.s.E(XY | \mathcal{G}) = X E (Y | \mathcal{G}) \text{ a.s.}

According to [1], χi\chi_{i} is Fi−1\mathcal{F}_{i-1}-measurable, and from the smoothing property and the definition of martingale, Yi−1Y_{i-1} is also Fi−1\mathcal{F}_{i-1}-measurable, therefore according to [3]: bEβN(0,b)≤∑i=1NE[E[(Yi−Yi−1)χi∣Fi−1]]≤∑i=1NE[χiE[(Yi−Yi−1)∣Fi−1]]=∑i=1NE[χiE[Yi∣Fi−1]−χiE[Yi−1∣Fi−1]]=∑i=1NE[χiE[Yi∣Fi−1]−χiYi−1] \begin{align*} b E \beta_{N} (0,b) \le & \sum_{i=1}^{N} E \left[ E \left[ ( Y_{i} - Y_{i-1}) \chi_{i} | \mathcal{F}_{i-1} \right] \right] \\ &\color{blue}{\le}& \sum_{i=1}^{N} E \left[ \chi_{i} E \left[ ( Y_{i} - Y_{i-1}) | \mathcal{F}_{i-1} \right] \right] \\ =& \sum_{i=1}^{N} E \left[ \chi_{i} E \left[ Y_{i} | \mathcal{F}_{i-1} \right] - \chi_{i} E \left[ Y_{i-1} | \mathcal{F}_{i-1} \right] \right] \\ &\color{red}{=}& \sum_{i=1}^{N} E \left[ \chi_{i} E \left[ Y_{i} | \mathcal{F}_{i-1} \right] - \chi_{i} Y_{i-1} \right] \end{align*} Since XnX_{n} was a submartingale by assumption, Yn=(Xn−a)+Y_{n} = ( X_{n} - a )^{+} is also a submartingale, hence E[Yi∣Fi−1]−Yi−1≥0E \left[ Y_{i} | \mathcal{F}_{i-1} \right] - Y_{i-1} \ge 0 holds. By eliminating indices that are χi=0\chi_{i} = 0 and only keeping those that are χi=1\chi_{i} = 1, the following inequality holds: bEβN(0,b)≤∑i=1NEχi[E[Yi∣Fi−1]−Yi−1]=∑i=1N[EE[Yi∣Fi−1]−EYi−1]=∑i=1N[EYi−EYi−1]=EYN−EY0 \begin{align*} b E \beta_{N} (0,b) \le & \sum_{i=1}^{N} E \chi_{i} \left[ E \left[ Y_{i} | \mathcal{F}_{i-1} \right] - Y_{i-1} \right] \\ =& \sum_{i=1}^{N} \left[ E E \left[ Y_{i} | \mathcal{F}_{i-1} \right] - E Y_{i-1} \right] \\ =& \sum_{i=1}^{N} \left[ E Y_{i} - E Y_{i-1} \right] \\ =& E Y_{N} - E Y_{0} \end{align*} Finally, since Yn=(Xn−a)+Y_{n} = ( X_{n} - a )^{+} was the case: EβN(a,b)≤EXN++∣a∣b−a E \beta_{N} (a,b) \le {{ E X_{N}^{+} + |a| } \over { b-a }}

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