1
$\begingroup$

Let $H \in (0,1)$ and let $\delta \ge 0 $ and let $d \in {\mathbb N}_+$. We are interested in computing a probability that, in a fractional Brownian Noise with Hurst exponent $H$ there are $n$ consecutive returns that are all bigger in magnitude than a threshold $\delta$.

In other words we want to find a closed form expression for the quantity below:

\begin{equation} {\mathcal J}^{(n)}(\delta,H):= \frac{1}{\sqrt{(2\pi)^n \det(\Sigma)}} \int\limits_{{\mathbb R}^n} e^{-\frac{1}{2} \vec{r} \cdot \Sigma^{-1} \cdot \vec{r}^T} \cdot \left(\prod\limits_{j=1}^d 1_{|r_j| \ge \delta } d r_j \right) \tag{1} \end{equation}

where $\Sigma:= \left( f^{(H)}(i-j)\right)_{i=1,j=1}^{n,n}$ with $f^{(H)}(l) := 1/2 \left( |1+l|^{2 H} - 2 |l|^{2 H} + |l-1|^{2 H}\right) $.

We managed to compute this quantity for $n=3$. We define auxiliary quantities as follows:

\begin{eqnarray} C_1(H) &:=& \frac{1}{\sqrt{\frac{\left(2^{2 H+1}-9^H-3\right) \left(-2^{6 H+1}+2\ 9^H+3\ 16^H-81^H+144^H-1\right)}{-2^{2 H+1}-9^H+16^H+1}}} \\ C_2(H) &:=&2^{-H} \sqrt{\frac{1}{4-4^H}} \\ D_1(H)&:=& -\frac{1}{\left(-2^{2 H+1}-9^H+16^H+1\right) \sqrt{\frac{1}{-2^{6 H+1}+2\ 9^H+3\ 16^H-81^H+144^H-1}}} \\ D_2(H)&:=& -\sqrt{\frac{1}{-2^{6 H+1}+2\ 9^H+3\ 16^H-81^H+144^H-1}} \end{eqnarray} and $ (l_j(H))_{j=1}^7 := \left\{2^{2 H+1}-9^H-1,4^H+2\ 9^H+16^H-36^H-2,-7\ 4^H+2\ 9^H+3\ 16^H-36^H-2,-2\ 9^H+16^H+2,3\ 4^H-9^H-5,4^H-9^H-1,4^H-2\right\}$.

With those definitions at hand our result reads as below:

\begin{align} \mathcal{J}^{(3)}(\delta, H) &= 3 - 3 \, \mathrm{erf}\left(\frac{\delta}{\sqrt{2}}\right) - 4\, T\left(\delta, \frac{2^{-H}}{\sqrt{\frac{1}{4 - 4^H}}}\right) - 4\, T\left(\delta, 2^H \sqrt{\frac{1}{4 - 4^H}}\right) \notag \\ &\quad + 4\, T\left(\delta, \frac{1}{(-3 + 2^{2H+1} - 9^H) C_1(H)}\right) - 2\, T\left(\delta, \frac{1}{(1 + 2^{2H+1} - 9^H) C_1(H)}\right) \notag \\ &\quad + 2\, T\left(\delta, (-3 + 2^{2H+1} - 9^H) C_1(H)\right) \notag \\ &\quad + \int_{\mathbb{R}} \bigg[ T\left(C_1(H)(2\delta + r l_1(H)), \frac{D_1(H)(\delta l_5(H) + r l_7(H))}{2\delta + r l_1(H)}\right) \notag \\ &\quad - T\left(C_1(H)(r l_1(H) - 2\delta), \frac{D_1(H)(\delta l_5(H) - r l_7(H))}{r l_1(H) - 2\delta}\right) \notag \\ &\quad + T\left(C_1(H)(2\delta + r l_1(H)), \frac{D_1(H)(\delta l_6(H) - r l_7(H))}{2\delta + r l_1(H)}\right) \notag \\ &\quad - T\left(C_1(H)(r l_1(H) - 2\delta), \frac{D_1(H)(\delta l_6(H) + r l_7(H))}{r l_1(H) - 2\delta}\right) \notag \\ &\quad + T\left(C_2(H)(2\delta + r l_7(H)), \frac{D_1(H)(\delta l_2(H) + r l_4(H))}{2\delta + r l_7(H)}\right) \notag \\ &\quad - T\left(C_2(H)(r l_7(H) - 2\delta), \frac{D_1(H)(\delta l_2(H) - r l_4(H))}{r l_7(H) - 2\delta}\right) \notag \\ &\quad + T\left(C_2(H)(r l_7(H) - 2\delta), \frac{D_1(H)(\delta l_3(H) - r l_4(H))}{r l_7(H) - 2\delta}\right) \notag \\ &\quad - T\left(C_2(H)(2\delta + r l_7(H)), \frac{D_1(H)(\delta l_3(H) + r l_4(H))}{2\delta + r l_7(H)}\right) \notag \\ &\quad + \frac{1}{4} \left(\mathrm{sgn}(4\delta - 2r l_1(H)) + \mathrm{sgn}(4\delta + 2r l_1(H))\right) \notag \\ &\qquad \times \left(\mathrm{sgn}(2\delta - r l_7(H)) + \mathrm{sgn}(2\delta + r l_7(H))\right) \bigg] \notag \\ &\quad \times \frac{e^{-\frac{1}{2} r^2}}{\sqrt{2\pi}} \, \mathbf{1}_{|r| \ge \delta} \, dr \qquad (2) \end{align}

The code below verifies the result.

(*Definitions*) Clear[CC1, DD1, DD2]; CC1[H_] := 1 / Sqrt[((-3 + 2^(1 + 2 H) - 9^H) (-1 - 2^(1 + 6 H) + 2 9^H + 3 16^H - 81^H + 144^H))/(1 - 2^(1 + 2 H) - 9^H + 16^H)]; CC2[H_] := 2^-H Sqrt[1/(4 - 4^H)]; DD1[H_] := -( 1/((1 - 2^(1 + 2 H) - 9^H + 16^H) Sqrt[( 1/(-1 - 2^(1 + 6 H) + 2 9^H + 3 16^H - 81^H + 144^H))] )); DD2[H_] := -Sqrt[( 1/(-1 - 2^(1 + 6 H) + 2 9^H + 3 16^H - 81^H + 144^H))] ; In[7342]:= (*Identity 4*) (*For a fractional Brownian Noise with drift what is the likelihood \ that three consecutive returns Subscript[r, k-l2,]Subscript[r, \ k-l1,]Subscript[r, k] are all bigger then a threshold? *) H = RandomReal[{0.45, 0.55}];(*Hurst exponent.*) dd = RandomReal[{0, 1}];(*Trigger*) {l1, l2} = RandomInteger[{1, 10}, 2];(*Lag values.*){l1, l2} = {1, 2}; b = RandomReal[{0, 1}, 3];(*External field.*)b = {0, 0, 0}; f[l_] := 1/2 (-2 Abs[l]^(2 H) + Abs[1 + l]^(2 H) + Abs[-1 + l]^(2 H)); Sig := {{1, f[l1], f[l2]}, {f[-l1], 1, f[l2 - l1]}, {f[-l2], f[l1 - l2], 1}}; AA = Inverse[Sig]; (*************) (*For a fractional Brownian Noise with drift what is the likelihood \ that three consecutive returns Subscript[r, k-l2,]Subscript[r, \ k-l1,]Subscript[r, k] are all bigger then a threshold? *) Print["H,dd,b,l1,l2=", {H, dd, b, l1, l2}]; 1/Sqrt[(2 \[Pi])^3 Det[Sig]] NIntegrate[ Exp[-1/2 {r1, r2, r3} . (AA . {r1, r2, r3})] If[ Abs[r1] > dd && Abs[r2] > dd && Abs[r3] > dd, 1, 0], {r1, -Infinity, Infinity}, {r2, -Infinity, Infinity}, {r3, -Infinity, Infinity}] (*Try to reduce the above to a closed form expression:*) Clear[l1, l2, l3, l4, l5, l6, l7]; l1[H_] := (-1 + 2^(1 + 2 H) - 9^H); l2[H_] := (-2 + 4^H + 2 9^H + 16^H - 36^H); l3[H_] := (-2 - 7 4^H + 2 9^H + 3 16^H - 36^H); l4[H_] := (2 - 2 9^H + 16^H); l5[H_] := (-5 + 3 4^H - 9^H); l6[H_] := (-1 + 4^H - 9^H); l7[H_] := (-2 + 4^H); (3 - 3 Erf[dd/Sqrt[2]] + -4 OwenT[dd, 2^-H/Sqrt[1/(4 - 4^H)]] - 4 OwenT[dd, 2^H Sqrt[1/(4 - 4^H)]] + 4 OwenT[dd, 1/((-3 + 2^(1 + 2 H) - 9^H) CC1[H] )] + 2 OwenT[dd, (-3 + 2^(1 + 2 H) - 9^H) CC1[H]] - 2 OwenT[dd, 1/((1 + 2^(1 + 2 H) - 9^H) CC1[H])] + 1/Sqrt[2 \[Pi]] NIntegrate[( ( -OwenT[CC1[H] (-2 dd + l1[H] r3), DD1[H] (l5[H] dd - l7[H] r3)/(-2 dd + l1[H] r3)] + OwenT[CC1[H] (2 dd + l1[H] r3), DD1[H] (l5[H] dd + l7[H] r3)/(2 dd + l1[H] r3)]) + (-OwenT[CC1[H] (-2 dd + l1[H] r3), DD1[H] (l6[H] dd + l7[H] r3)/(-2 dd + l1[H] r3)] + OwenT[CC1[H] (2 dd + l1[H] r3), DD1[H] (l6[H] dd - l7[H] r3)/(2 dd + l1[H] r3)]) + (-OwenT[CC2[H] (-2 dd + l7[H] r3), DD2[H] (l2[H] dd - l4[H] r3)/(-2 dd + l7[H] r3)] + OwenT[CC2[H] (2 dd + l7[H] r3), DD2[H] (l2[H] dd + l4[H] r3)/(2 dd + l7[H] r3)]) + (OwenT[CC2[H] (-2 dd + l7[H] r3), DD2[H] (l3[H] dd - l4[H] r3)/(-2 dd + l7[H] r3)] - OwenT[CC2[H] (2 dd + l7[H] r3), DD2[H] (l3[H] dd + l4[H] r3)/(2 dd + l7[H] r3)]) + ( (Sign[2 dd - l7[H] r3] + Sign[2 dd + l7[H] r3]) (Sign[4 dd + 2 l1[H] r3] + Sign[4 dd - 2 l1[H] r3]))/4 ) E^(-(r3^2/2)) If[Abs[r3] > dd, 1, 0], {r3, -Infinity, Infinity}] ) During evaluation of In[7342]:= H,dd,b,l1,l2={0.486717,0.768615,{0,0,0},1,2} Out[7350]= 0.0864547 Out[7359]= 0.0864547

Now, my question is the following. Is is possible to express the the last integral in the right hand side in (2) through some known special functions? In other words we ask to find the antiderivative $ \int \exp(-1/2 x^2) T(a x+ b, \frac{c_1 x+ d_1}{c_2 x+ d_2}) dx = ?$

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

We define a generalized Owen T function as follows:

\begin{equation} T(h,a,K^{(x)}) := \frac{1}{2 \pi} \int_0^a \frac{\exp(-\frac{1}{2} h^2(1+\eta^2))}{1+\eta^2} \cdot K^{(x)}(\eta) d\eta \tag{1} \end{equation}

The function in question clearly reduces to the Owen's T function if the kernel $K^{(x)}$ is identically equal to unity.

Now, it turns out, that for some specific choice of the kernel the function (1) is the anti-derivative being sought after.

To be specific let us assume that $x>0$ then $a,b \in {\mathbb R}$ and $c_1,c_1,d_1,d_2>0$ . We also define a kernel \begin{eqnarray} &&K^{(x;a,b,c_1,c_2,d_1,d_2)}(\eta) := \\ &&\frac{\sqrt{\frac{\pi }{2}} e^{\frac{a^2 b^2 \left(\eta ^2+1\right)^2}{2 a^2 \left(\eta ^2+1\right)+2}} } { \sqrt{a^2 \left(\eta ^2+1\right)+1} } \cdot \\ && % \left\{ \begin{array}{lll} {\bar K} (\eta, 1_{\eta < \frac{c_1 x+ d_1}{c_2 x+d_2}} x + (1-1_{\eta < \frac{c_1 x+ d_1}{c_2 x+d_2}}) \frac{-d_1+d_2 \eta}{c_1-c_2 \eta} ) - {\bar K}(\eta,0) & if & c_1 d_1- c_1 d_2 > 0 \\ % {\bar K}(\eta,x) - {\bar K}(\eta,1_{\eta < \frac{d_1}{d_2}} \cdot 0 + (1-1_{\eta < \frac{d_1}{d_2}}) \frac{-d_1+d_2 \eta}{c_1-c_2 \eta}) & if & c_1 d_1- c_1 d_2 < 0 \\ \end{array} \right. \end{eqnarray}

Then we have:

\begin{eqnarray} &&\int\limits_0^x \exp(-\frac{1}{2} \xi^2) T(a \xi+b, \frac{c_1 \xi + d_1}{c_2 \xi + d_2}) d\xi = \\ &&T(b,\frac{d_1}{d_2} \cdot 1_{c_1 d_1- c_1 d_2 > 0} + \frac{c_1 x+ d_1}{c_2 x+ d_2} \cdot 1_{c_1 d_1- c_1 d_2 < 0} ,K^{(x;a,b,c_1,c_2,d_1,d_2)}) \end{eqnarray}

In[9442]:= Clear[KK]; KK[eta_, xi_] := Erf[(a b (1 + eta^2) + (1 + a^2 (1 + eta^2)) xi)/Sqrt[ 2 + 2 a^2 (1 + eta^2)]]; (*(c1 xi+d1)/(c2 xi+d2)==c1/c2+(c2 d1-c1 d2)/(c2 (d2+c2 xi))*) x = RandomReal[{0, 2}]; {a, b} = RandomReal[{-2, 2}, 2]; {c1, c2} = RandomReal[{0, 2}, 2]; {d1, d2} = RandomReal[{0, 2}, 2]; NIntegrate[ Exp[-1/2 xi^2] OwenT[a xi + b, (c1 xi + d1)/(c2 xi + d2)], {xi, 0, x}] If[Sign[c2 d1 - c1 d2] == 1, 1/(2 Pi) NIntegrate[ Exp[-1/2 b^2 (1 + eta^2)]/(1 + eta^2) ((E^((( a^2 b^2 (1 + eta^2)^2)/(2 + 2 a^2 (1 + eta^2))))) Sqrt[(\[Pi]/ 2)] )/Sqrt[ 1 + a^2 (1 + eta^2)] (KK[eta, If[eta <= (c1 x + d1)/(c2 x + d2), x, (-d1 + d2 eta)/( c1 - c2 eta)]] - KK[eta, 0]), {eta, 0, d1/d2}], 1/(2 Pi) NIntegrate[ Exp[-1/2 b^2 (1 + eta^2)]/(1 + eta^2) ((E^((( a^2 b^2 (1 + eta^2)^2)/(2 + 2 a^2 (1 + eta^2))))) Sqrt[(\[Pi]/ 2)] )/Sqrt[ 1 + a^2 (1 + eta^2)] (KK[eta, x] - KK[eta, If[eta <= (d1)/(d2), 0, (-d1 + d2 eta)/( c1 - c2 eta)]]), {eta, 0, (c1 x + d1)/(c2 x + d2)}] ] Out[9448]= 0.0753323 Out[9449]= 0.0753323
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.