valami

$$\int f_{\alpha }(t)dt=K(f_{\alpha }(\cdot ))$$ $$f_{\alpha }(t)=E{\left|S_t\right|}^{\frac{\alpha }{\alpha -1}}$$ $$Q(T):={\int }_0^{T}E{\left|S_t\right|}^{\frac{\alpha }{\alpha -1}}dt$$ $$\tilde{Q}(T):=\mathbb{E}\left({\int }_0^T{\left|S_t\right|}^{\frac{\alpha }{\alpha -1}}dt\right)$$ $$\bar{Q}(T):=\mathbb{E}{\left({\int }_0^T\left|S_t\right|dt\right)}^{\frac{\alpha }{\alpha -1}}$$ $$f_{\alpha }(t)\approx$$ $$S_t\sim N(0,t^{2H})$$

• aszimptotikosan optimalis strategianak a konstansszorzojat keressuk meg binaris keresessel
• $Q$ market bound-ot tuntessuk fel a train-eval ploton
• kenyszeritsuk az agentet likvidalasra

Hogyan szamoljuk ki a market bound-ot, $Q$-t?
Ebben a cikkben talalhato a kovetkezo formula:

$$\mathbb{E}[|X{|}^{\nu }]={\sigma }^{\nu }{2}^{\nu /2}\frac{\Gamma \left(\frac{\nu +1}{2}\right)}{\sqrt{\pi }}\cdot \Phi \left(-\frac{\nu }{2},\frac{1}{2},-\frac{{\mu }^2}{2{\sigma }^2}\right)$$

Mivel az FBM-nek minden idopontban a varhatoerteke nulla, $\mu =0$, ezert a kovetkezore egyszerusul a fenti:

$$\mathbb{E}[|X{|}^{\nu }]={\sigma }^{\nu }{2}^{\nu /2}\frac{\Gamma \left(\frac{\nu +1}{2}\right)}{\sqrt{\pi }}$$

Tehat mar csak a szorast kell kiszamolni.

A wikipedia szocikkben talalhato, hogy a kovariencia fuggvenye az FBM-nek a kovetkezo:

$$\mathbb{E}\left[B_H(t)B_H(s)\right]=\frac{1}{2}\left(|t{|}^{2H}+|s{|}^{2H}-|t-s{|}^{2H}\right)$$

Ha $t=s$ akkor pont a szorast kapjuk meg, mert mindenhol nulla a varhatoertek

$${\sigma }^2(t)=\mathbb{E}[(B_H(t)-\mathbb{E}[B_H(t)]{)}^2]=\mathbb{E}[B_H(t{)}^2]=\frac{1}{2}\left(|t{|}^{2H}+|t{|}^{2H}-|t-t{|}^{2H}\right)$$

Egyszerusitesek utan a kovetkezot kapjuk:

$${\sigma }^2(t)=|t{|}^{2H}$$

Visszaterve a market bound-ra, a kovetkezo objektumot szeretnenk meg mindig kiszamolni, amihez mar meg van az osszes eszkozunk.

$$Q(T):={\int }_0^{T}E{\left|S_t\right|}^{\frac{\alpha }{\alpha -1}}dt$$ $$Q(T)=\underset{0}{\overset{T}{\int }}{\sigma }^{\alpha /(\alpha -1)}{2}^{\alpha /2(\alpha -1)}\frac{\Gamma \left(\frac{\alpha }{2(\alpha -1)}+\frac{1}{2}\right)}{\sqrt{\pi }}dt$$

Behelyettesitve ${\sigma }^2(t)=|t{|}^{2H}$ erteket:

$$Q(T)=\underset{0}{\overset{T}{\int }}|t{|}^{2H\cdot \alpha /(\alpha -1)}{2}^{\alpha /2(\alpha -1)}\frac{\Gamma \left(\frac{\alpha }{2(\alpha -1)}+\frac{1}{2}\right)}{\sqrt{\pi }}dt$$

Legyen $\beta :=\frac{\alpha }{\alpha -1}$

$$Q(T)=\underset{0}{\overset{T}{\int }}|t{|}^{2H\beta }{2}^{\beta }\frac{\Gamma \left(\frac{\beta +1}{2}\right)}{\sqrt{\pi }}dt$$ $$Q(T)=2^{\beta }\frac{\Gamma \left(\frac{\beta +1}{2}\right)}{\sqrt{\pi }}\underset{0}{\overset{T}{\int }}|t{|}^{2H\beta }dt$$ $$\sum _{u=1}^{T+1}{\Phi }_u\left(S_u-S_{u-1}\right)=-\sum _{u=0}^T{\varphi }_u{S}_u+S_{T+1}\sum _{u=0}^T{\varphi }_u$$ $$-\sum _{u=0}^T{\varphi }_u{S}_u+\sum _{u=0}^T{S}_u{\varphi }_u$$ $${\Phi }_{u+1}-{\Phi }_u:={\varphi }_u$$ $${\Phi }_{t+1}-{\Phi }_t={\varphi }_t$$ $$G^{\ast }(y):=\underset{x\in \mathbb{R}}{\sup }(xy-G(x))=\frac{\alpha -1}{\alpha }{\alpha }^{\frac{1}{1-\alpha }}{\lambda }^{\frac{1}{1-\alpha }}|y{|}^{\frac{\alpha }{\alpha -1}}$$ $$\left|X_T^0(\varphi )\right|\le \left|{\int }_0^T{G}^{\ast }\left(-S_t\right)dt\right|\le C{\int }_0^T{\left|S_t\right|}^{\alpha /(\alpha -1)}dt$$ $$\left|{\int }_0^T{G}^{\ast }\left(-S_t\right)dt\right|=\left|{\int }_0^T\frac{\alpha -1}{\alpha }{\alpha }^{\frac{1}{1-\alpha }}{\lambda }^{\frac{1}{1-\alpha }}|-S_t{|}^{\frac{\alpha }{\alpha -1}}dt\right|=$$ $$\gamma =\frac{1}{1-\alpha }$$ $$\left|{\int }_0^T{G}^{\ast }\left(-S_t\right)dt\right|=\frac{1}{\beta }{\alpha }^{\gamma }{\lambda }^{\gamma }\cdot \left|{\int }_0^T|-S_t{|}^{\frac{\alpha }{\alpha -1}}dt\right|$$