Create MarketBackFinal 2.0

[OrsonTyphanel93]

Backdoor attack via jumps-Diffusion and stock market and Bayesian Optimization : MarketBackFinal 2.0

Description

Hi guys @beat-buesser !, I am attaching the latest backdoor attack method based on Dynamic Hedging and rough volatility paths, the attack method is called MarketBackFinal 2.0 and it focuses explicitly on Bayesian optimization (BO) of trading processes. jump via a stochastic diffusion method.

Testing

MarketBackFinal 2.0 : For Bayesian optimization, an advanced black-box objective function is used in order to be able to use quadratic terms introducing non-linearity, sinusoidal components which add periodic fluctuations, making the oscillatory function more complex, thus introducing multiple local minima and maxima:

$$f(x, y) = -(x^{2} + y^{2} + 0.1xy + \sin{x} + \cos{y})$$

Update September 12rd, 2024 , last notebook,link

This is the last notebook, the next step will be to incorporate methods into the ART.1.19 framework.

Last update September 20, 2024 , last notebook,link

Reference, Figure

[update notebook MarketBackFinal 2.0 ( best version for fluid comprehension ) link..., notebook link(This notebook corresponds to the two figures you are observing )](https://github.com/OrsonTyphanel93/adversarial-robustness-toolbox/blob/dev_1.14.0/Udpdate_ART_1_18_MarketBackFinal_2_0_adversarial_machine_learning.ipynb

notebook link(This notebook corresponds to all the other figures you'll see, the one where I didn't include the vega and the rho on the studs.)

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The pathwise estimator of the option delta is $$\frac{dY}{dS(0)} = e^{-rT} 1_{\bar{S} > K} \frac{\bar{S}}{S(0)}.$$

$$S(t_i) = S(t_{i-1})e^{(r - \frac{1}{2}\sigma^2)(t_i - t_{i-1})} + \sigma\sqrt{t_i - t_{i-1}}Z_i, \quad i = 1, \dots, m,$$

with $Z_1,\ldots,Z_m$ independent standard normal random variables. The parameter $\sigma$ affects $S(t_i)$ explicitly through this functional relation but also implicitly through the dependence of $S(t_{i-1})$ on $\sigma$. By differentiating both sides, we get a recursion for the derivatives along the path:

$$\frac{dS(t_i)}{d\sigma} = \frac{dS(t_{i-1})}{d\sigma} \frac{S(t_i)}{S(t_{i-1})} + S(t_i)\left[-\sigma(t_i - t_{i-1}) + \sqrt{t_i - t_{i-1}}Z_i\right],$$

With initial condition $\frac{dS(0)}{d\sigma}=0$, this recursion is solved by

$$\frac{dS(t_i)}{d\sigma} = S(t_i)\left[\log\left(\frac{S(t_i)}{S(0)}\right) - (r + \frac{1}{2}\sigma^2)t_i\right]/\sigma.$$

The pathwise estimator of the Asian option vega is

$$e^{-rT} \frac{1}{m} \sum_{i=1}^{m} \frac{dS(t_i)}{d\sigma} 1_{\bar{S} > K}.$$

$$\begin{aligned} & \mathbb{E}\left[e^{-r T} V_T-e^{-r T}\left(S_T-K\right)_{+}\right]^2 \ &= {\left[V_0-v\left(0, S_0\right)\right]^2+\mathbb{E}\left(\int_0^T e^{-2 r s} S_{s-}^2 \sigma^2\left[v_x^{\prime}\left(s, S_{s-}\right)-\delta(s)\right]^2 d s\right.} \ &\left.\quad+\int_0^T \lambda \int_{\mathbb{R}} e^{-2 r s}\left[v\left(s, S_{s-}(1+y)\right)-v\left(s, S_{s-}\right)-\delta(s) y S_{s-}\right]^2 \nu(d y) d s\right) \end{aligned}$$

$$\begin{aligned} & e^{-r T}\left(S_T-K\right)_{+} \ & =v\left(0, S_0\right)+\int_0^T e^{-r t}\left(v_t^{\prime}\left(t, S_{t-}\right)+\mu S_{t-} v_x^{\prime}\left(t, S_{t-}\right)+\frac{1}{2} \sigma^2 S_{t-}^2 v_{x x}^{\prime \prime}\left(t, S_{t-}\right)-r v\left(t, S_{t-}\right)\right) d t \ & +\int_0^T e^{-r t} \sigma S_{t-} v_x^{\prime}\left(t, S_{t-}\right) d W_t+\int_{10, T]} e^{-r t}\left(v\left(t, S_{t-}\left(1+Y_t\right)\right)-v\left(t, S_{t-}\right)\right) d N_t \ & =v\left(0, S_0\right)+\int_0^T e^{-r t} \sigma S_{t-} v_x^{\prime}\left(t, S_{t-}\right) d W_t+\int_{\mid 0, T]} e^{-r t}\left(v\left(t, S_{t-}\left(1+Y_t\right)\right)-v\left(t, S_{t-}\right)\right) d N_t \ & \quad-\int_0^T e^{-r t} \lambda \int_{\mathbb{R}}\left(v\left(t, S_{t-}(1+y)\right)-v\left(t, S_{t-}\right)\right) \nu(d y) d t . \end{aligned}$$

$$\begin{aligned} e^{-r T} V_T & =V_0+\int_0^T \delta(t) e^{-r t} S_{t-}\left((\mu-r) d t+\sigma d W_t\right)+\int_{[0, T]} \delta(t) e^{-r t} S_{t-} Y_t d N_t \ & =V_0+\int_0^T e^{-r t} \sigma S_{t-} \delta(t) d W_t \ & +\int_{10, T]} \delta(t) e^{-r t} S_{t-} Y_t d N_t-\int_0^T \lambda \delta(t) e^{-r t} S_{t-} \int_{\mathbb{R}} y \nu(dy) dt \end{aligned}$$

Let $V_t^C$ be the value of a Call option at time $t$ whose payoff is $\Lambda^C$ on the underlying $S(t)$

$$V_t^C=\mathbb{E}^{\mathbb{Q}}\left[e^{-r(T-t)} \Lambda^C \mid \mathcal{F}_t\right]$$

$$ \begin{aligned} V_t^C= & S(t) \Phi\left(\frac{\ln \frac{S(t)}{K}+\left(r+\sigma^2 / 2\right)(T-t)}{\sigma \sqrt{T-t}}\right) \ & -K e^{-r(T-t)} \Phi\left(\frac{\ln \frac{S(t)}{K}+\left(r-\sigma^2 / 2\right)(T-t)}{\sigma \sqrt{T-t}}\right) \ V_t^C= & S(t) \Phi\left(d_{1, t}\right)-K e^{-r(T-t)} \Phi\left(d_{2, t}\right) \end{aligned} $$

where $\Phi$ is the cumulative distribution function of a standard normal random variable, and $d_{1, t}$ and $d_{2, t}$ are given as follows:

$$ d_{1, t}=\frac{\ln \frac{S(t)}{K}+\left(r+\sigma^2 / 2\right)(T-t)}{\sigma \sqrt{T-t}} $$

and

$$ \begin{aligned} d_{2, t} & =\frac{\ln \frac{S(t)}{K}+\left(r-\sigma^2 / 2\right)(T-t)}{\sigma \sqrt{T-t}} \ & =d_{1, t}-\sigma \sqrt{T-t} \end{aligned} $$

Then the unique rational price at time 0 of the Call option can be found betting $t=0$ in the equations of $V_t^C, d_{1, t}$ and $d_{2, t}$ to get the famous Black-Scholes pricing formula:

$$V_0^C=S(0) \Phi\left(d_{1,0}\right)-K e^{-r T} \Phi\left(d_{2,0}\right)$$

Fourier transform method for price calculation,sources

Reference 1 Reference 2 Reference 3

FAQ

to quantify the measure of financial risk, you can apply in the code of this notebook dating from August 2024, this code attached in this notebook.

We quantify the simulations by determining the potential financial risks that may arise over a given time horizon.

A zero coupon bond price subject to default risk (at time $t$ ) with a tenor of $T$ years can be defined as follows ,

$$\tilde{\delta}(t, T)=\underbrace{\mathbb{E}_t^{\mathbb{Q}}\left(e^{-\int_t^T r(u) d u} \mathbb{I}_{\tau>T}\right)}_{}+\underbrace{\mathbb{E}_t^{\mathbb{Q}}\left(e^{-\int_t^T r(u) d u} R \mathbb{I}_{\tau \leq T}\right)}_{}$$

where $R$ denotes the recovery rate and $r(t)$ is the instantaneous short-rate process. There is a single unit cash-flow at time $T$, the pay-off is $R \in[0,1]$.

$$\begin{aligned} \tilde{\delta}(t, T) & =\mathbb{E}_t^{\mathbb{Q}}\left(e^{-\int_t^T r(u) d u}\right) \mathbb{E}_t^{\mathbb{Q}}\left(\mathbb{I}_{\tau>T}\right)+\mathbb{E}_t^{\mathbb{Q}}\left(e^{-\int_t^T r(u) d u}\right) R \mathbb{E}_t^{\mathbb{Q}}\left(\mathbb{I}_{\tau \leq T}\right) \ & =\delta(t, T) \mathbb{Q}(\tau>T)+\delta(t, T) R \mathbb{Q}(\tau \leq T) \ & =\delta(t, T)(\mathbb{Q}(\tau>T)+R \mathbb{Q}(\tau \leq T)) \end{aligned}$$

where $\delta(t, T)$ and $\tilde{\delta}(t, T)$ represent risk-free and default-risky $T$ period zero-coupon bonds, respectively. The terms $\mathbb{Q}(\tau>T)$ and $\mathbb{Q}(\tau \leq T)$ .

Usage/Examples mathematics concept see

   
    def tail_probabilities_and_shortfall_integrals(self, S, K, T, r, sigma, threshold=5):
        """
        Calculate tail probabilities and shortfall integrals for a call option.

         Parameters:
         S (float): Underlying asset price
         K (float): Strike price
         T (float): Time to expiration
         r (float): Risk-free interest rate
         sigma (float): Volatility of the underlying asset
         threshold (float): Threshold value for shortfall calculation

         Returns:
        tuple: (tail_probability, shortfall_integral)
        """
        # Calculate d1 and d2
        d1 = (np.log(S / K) + (r + 0.5 * sigma**2) * T) / (sigma * np.sqrt(T))
        d2 = d1 - sigma * np.sqrt(T)

        # Calculate tail probability
        tail_probability = 1 - norm.cdf(d2)

        # Calculate shortfall integral
        shortfall_integral = (K * np.exp(-r * T) * norm.cdf(-d2) - S * norm.cdf(-d1)) - threshold

        # Handle potential numerical issues
        tail_probability = np.clip(tail_probability, 0, 1)
        shortfall_integral = np.clip(shortfall_integral, 0, None)


        
        # Calculate VaR for 95% and 99% confidence levels
        var_95 = norm.ppf(0.05, d1, sigma * np.sqrt(T))
        var_99 = norm.ppf(0.01, d1, sigma * np.sqrt(T))

        # Convert VaR values to percentage loss
        var_95_percent = (1 - np.exp(var_95)) * 100
        var_99_percent = (1 - np.exp(var_99)) * 100

        return tail_probability, shortfall_integral, var_95_percent, var_99_percent

     # Then apply exactly this in the Bayesian function only, and you'll get all the numerical values 
     tail_prob, shortfall_int,var_95_percent, var_99_percent = self.tail_probabilities_and_shortfall_integrals(S, K, T,                r,sigma, threshold=5)
       print(f"Tail Probability: {tail_prob}")
       print(f"Shortfall Integral: {shortfall_int}")
       print(f"95% VaR: {var_95_percent}")
       print(f"99% VaR: {var_99_percent}")

Usage/Examples McKean-Vlasov process

   
    def mean_field_interaction(self,x):
        """
        Compute the mean field interaction for the McKean-Vlasov process.

        Parameters:
        x (numpy array): Current state of the process

        Returns:
        float: Mean field interaction value
        """
        # Use np.mean() for efficiency
        mean_field_effect = np.mean(x)

        # Add a small constant to avoid division by zero
        epsilon = 1e-8

        # Compute the squared difference between each element and the mean
        squared_diff = (x - mean_field_effect)**2

        # Compute the mean of the squared differences
        var_x = np.mean(squared_diff)

        # Compute the second-order correction term
        second_order_correction = 0.5 * var_x


        # Combine terms
        drift_term = mean_field_effect + second_order_correction

        return drift_term


    def mckean_vlasov_process(self,P0, mu, sigma, t_max, dt, num_steps):
        """
        Simulate the McKean-Vlasov process.

        Parameters:
        S0 (float): Initial value of the process
        mu (callable): Drift term function
        sigma (float): Volatility
        t_max (float): Maximum time
        dt (float): Time step size
        num_steps (int): Number of steps

        Returns:
        numpy array: Array of simulated values at each time step
        """
        n = int(t_max / dt)
        t = np.linspace(0, t_max, n+1)
        # Initialize the process
        P = np.zeros(n+1)
        P[0] = P0

        # Generate random numbers for the Brownian motion
        # Simulate the process
        for i in range(1, n+1):
            dW = norm.rvs(size=1)
            P[i] = P[i-1] * np.exp((mu(P[i-1]) - 0.5 * sigma**2) * dt + sigma * np.sqrt(dt) * dW)


        return t, P

        t, P = self.mckean_vlasov_process(P0=1.0, mu=self.mean_field_interaction, sigma=0.5, t_max=1.0, dt=0.01, num_steps=10000)
        print(f"Mckean-Vlasov Process: Time={t}, Value={S}")


        plt.figure(figsize=(12, 6))
        plt.plot(t, P)
        plt.title("McKean-Vlasov Process Simulation")
        plt.xlabel("Time")
        plt.ylabel("Value")
        plt.grid(True)
        plt.savefig('mckean_vlasov_process_simulation.png', dpi=300, bbox_inches='tight')
        plt.show()