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5

It is, of course, possible to price such a contract in a no-arbitrage market. Indeed, if $f$ is a sufficiently smooth function, then you can price all contracts paying $f(S_T)$. Note that your specific payoff has no optionality and that the payoff may be negative. Bakshi and Madan (2000) discuss the economic meaning of a derivative paying $\cos(S_T)$ in the ...


2

Your solution is correct. Rewriting your modified payoff in terms of the payoff of a call option is a common technique. Note however that you have a little typo: you need $S(t)e^{-\delta(T-t)}$ instead of $S(T)$ in the last line, i.e. \begin{align*} \Pi(t,X) &= 1+S_te^{-\delta(T-t)}N(d_1)-N(d_2) \\ &= S_te^{-\delta(T-t)}N(d_1) +N(-d_2), \end{align*} ...


3

Yes you can use the Black-Scholes Model with $K=B(T)$ because $B(T)$ is deterministic (a constant like $K$, since $T$ is constant). However your current solution is incorrect as the Black-Scholes call price is already discounted ($C:=e^{-rT}E^Q[(S_T-K)^+])$ and based on the current stock price $S_t$ (not $S_T$). Further note that $e^{-r(T-t)}B(T)=1$ and $1-...


1

Antithetic Sampling is used to reduce variance in simulation by sampling from two 'opposite' set of distributions. Since in Heston models you require sampling from normal for both the price and volatility, it would be better to use antithetic sampling for both.


0

Not sure I understand well your question, but in a stochastic volatility model you have to implement your discretised process for both underlying and volatility since the underlying's move are conditional on the volatility process value. Note that more efficient discretisation schemes as Euler exist for the CIR process followed by variance in Heston's model ...


8

$\max(B_T,S_T)=\max(0,S_T-B_T)+B_T,$ so this is just a call option (with strike $B_T$) plus $B_T.$


2

You have $$dZ_t = df\left(S_t, B_t, X_t\right) = \frac{\partial f}{\partial s}dS_t + \frac{\partial f}{\partial b}dB_t + \frac{\partial f}{\partial x}dX_t + \frac{1}{2}\left[\frac{\partial^2 f}{\partial s^2} d\langle S\rangle_t + 2\frac{\partial^2 f}{\partial s\partial x} d\langle S, X\rangle_t \right]$$ since $d\langle B\rangle_t$, $d\langle B, S\rangle_t$, ...


1

The idea is to use Fubini's theorem to interchange expectations with respect to the Brownian path with the integral. Thus $\mathbb EX_t=\int_0^t\mathbb EW_t\ dt=0$ and $$ \mathbb E(X_t^2)=\mathbb E\int_0^t\int_0^t W_uW_v\ dv \ du=\int_0^t\int_0^t \mathbb E(W_uW_v)\ dv\ du=\int_0^t\int_0^t\min(u,v)\ dv\ du, $$ using the covariance of the Brownian motion in ...


4

If you are happy to try the brute force approach, then here are the relevant formulae: In ordinary calculus, you have the product rule for the differential of two variables: $$d \left( x_1 x_2\right)=x_1 dx_2+x_2 dx_1$$ The general version of this for differential of products of n variables is: $$d \prod_{i=1}^{n}{x_i}=\sum_{i=1}^{n}{ \prod_{j=1 j \ne i}^...


7

Physical objects move according to simple smooth curves that can be represented by low order polynomials: a straight line, a parabola, an ellipse, etc. Financial market prices move in a completely different way, as can be seen by looking at any graph of stock prices, interest rates etc. in a newspaper: there are constant, erratic fluctuations, sometimes in ...


15

Brownian motion is simply the limit of a scaled (discrete-time) random walk and thus a natural candidate to use. It is very intuitive and arguably one of the simplest and best understood time-continuous stochastic processes. Also, don't forget that you obtain many more stochastic processes as functions of a (time-changed) Brownian motion. In many books on ...


3

For the mean you can use Fubini's Theorem to change the order of integration $$ E\int_0^t (W_s)^n ds = \int_0^t E(W_s)^n ds$$ Then we can use the fact that $W_s \sim N(0,\sqrt{s})$ to obtain \begin{align*} E (W_s)^{2k+1} &= 0, k=0,1,2,... \tag*{(odd n)} \\ E (W_s)^{2k} &= (2k-1)!!s^k, k=1,2,... \tag*{(even n)} \end{align*} Therefore, \begin{align*...


2

If $S_T<K_1$, the payoff is zero, and we have $\frac{(K_2-K_1)S(T)}{K_2} \geq0$ If $K_1 \leq S_T<K_2$, the payoff is $(S_T -K_1)$. We have $$K_1K_2 \geq S_TK_1$$ and $$S_TK_2+K_1K_2 \geq S_TK_2+S_TK_1$$ Thus, $$S_TK_2-S_TK_1 \geq S_TK_2-K_1K_2$$ Finally, $$S_T(K_2-K_1) \geq (S_T-K_1)K_2$$ $$\frac{S_T(K_2-K_1)}{K_2} \geq (S_T-K_1)$$ If $S_T \geq ...


3

Shreve is a bit naughty here but, of course, he is right. When you have the risk-neutral measure $\mathbb{Q}$ or $\tilde{\mathbb{P}}$, you can price derivatives as discounted expectation by the very definition of the risk-neutral measure (better called: equivalent martingale measure). So indeed, once you have $\tilde{\mathbb{P}}$, you can price derivatives ...


2

For Risk-neutral Pricing to “work”, you need assumptions where risk elimination by trading financial instruments is possible : no counterparty risk, no transaction costs, continuous trading, continuous asset paths. If such assumptions are not fulfilled (which is the case in real markets ; however, for large banks they are sufficiently near from reality), ...


4

To prove that $Y(T)=0$, you integrate the SDE : $$\int_{0}^{T}dY(t) = \int_{0}^{T}\nu dW(t) - \frac{\nu}{T} \int_{0}^{T}W(T) dt$$ $$Y(T)-Y(0)=\nu\left( W(T)-W(0)\right)- \frac{\nu}{T}W(T)\left(T-0\right)$$ Therefore $$Y(T)=Y(0)$$ You forgot to write the initial conditions but I guess $Y(0)=0.$ EDIT : The dependency in $\nu$ is irrelevant in pricing ...


2

The confusion is that you think that we define the numeraire as this exponential function... It is not the case. We give the numeraire properties to $N$, then we model it. Similar to any other model. All we know is that $N$ is positive, and we have $$\frac{V_t}{N_t}=E^{N}\left[\frac{V_T}{N_T}|\mathbb{F_t}\right]$$ where $V_t$ is a tradable asset. $N$ can ...


4

In simple terms, more time to expiry $T$ increases the value of an at-the-money (ATM) option as it gives more time for the stock to rise further (or fall further in the case of a put option). This means that the potential upside of the option is greater (the downside is not as it is floored at zero). So the option is worth more. This effect has nothing to do ...


1

This is rather similar to the solution you mentioned in your question :) Let $(r_t)$ be the short rate with $\int_0^{t}r_s\mathrm{d}s\sim N(0.03t,0.25t)$ and $B_t$ the value of the bank account. Recall that by definition $\mathrm{d}B_t=r_tB_t\mathrm{d}t$ and thus $B_t=B_0\exp\left(\int_0^t r_s\mathrm{d}s\right)$. Thus, $(B_t)$ is for every time point $t$ ...


2

Here, we use the time-changed Brownian motion technique to show the normality of \begin{align*} Y_t = \int_0^t u\, dW_u, \end{align*} where $\{W_t, \, t \ge 0\}$ is a standard Brownian motion with respect to the filtration $\{\mathscr{F}_t,\, t \ge 0\}$. For $t\ge 0$, let $\mathscr{G}_t = \mathscr{F}_{\sqrt[3]{3t}}$. Consider the process $M=\{M_t, \, t\ge 0\...


2

It is better to express $X$ as $X_t = \frac{1}{t} \int_{0}^{t} u \, d W_u$. The mean of $X$ is given by $$ \mathbb{E}[X_t]=\frac{1}{t} \mathbb{E} \left[ \int_{0}^{t} u \, d W_u \right] = \frac{1}{t} 0 = 0 $$ and the variance of $X$ is given by $$ \mathbb{E}[X^2_t]=\frac{1}{t^2} \mathbb{E} \left[ \left( \int_{0}^{t} u \, d W_u \right)^2 \right] = \frac{1}{t^2}...


4

That is not the SDE for the Heston model - it violates the affine property in the drift term. In other words, the paper has a typo. The correct SDE is: $$ d v_t = \kappa (m-v_t) dt + \nu \sqrt{v_t} dw_t $$ where $v_t := \sigma_t^2$ is the variance. Let $\xi_t^T := \mathbb{E}_t [ v_T]$ denote the forward variance and see that $$ \begin{align} \xi_{t}^{T} &...


5

I think what is meant is that the increment of the brownian between discrete points $(i-1)$ and $i$ is normally distributed with mean 0 and variance equal to $\delta t$ which represents the length of the interval between the two discrete points. You can then write the increment as $\sqrt{\delta t}$ times a standard normal random. So the equation is to be ...


2

The integral $I_T$ is an Itô stochastic integral therefore its expectation is $0$. This is because $I_T$ is a martingale (see e.g. Theorem 4.3.1 in Shreve), hence: $$\mathbb{E}[I_T]=I_0=0$$ You can also see this by considering the definition of a stochastic integral, which involves the sum of terms of the form $f(W_{t_i})(W_{t_{i+1}}-W_{t_i})$, and using the ...


1

It is easier if you interpret t as time in years. So let’s say t=4 years. And the rest is easier if you recall the end result, we want this scaled random walk to approach the standard brownian, which has mean zero and variance t (seeing it as interval from time 0 to t=4). We are repeating independent and identical tossing game, where the coin is unbiased. ...


2

Note that, for $t>s>0$, \begin{align*} X_t-X_s &= \frac{1}{t}\int_0^t udW_u - \frac{1}{s}\int_0^s udW_u\\ &=\frac{1}{t}\bigg(\int_s^t u dW_u + \int_0^s udW_u \bigg)- \frac{1}{s}\int_0^s udW_u\\ &=\frac{1}{t} \int_s^t u dW_u + \Big(\frac{1}{t} -\frac{1}{s}\Big)\int_0^s udW_u\\ &=\frac{1}{t} \int_s^t u dW_u - \frac{t-s}{t} X_s. \end{align*...


1

If we denote the random walk with $(X_k)_{k \in \mathbb{N}}$ than for all $k$ the random variable $\Delta X_k := X_{k} - X_{k-1}$ has mean zero and variance one: \begin{align} \mathbb{E}[\Delta X_k] = \frac 12\cdot 1 + \frac 12 \cdot (-1) = 0, \quad \text{Var}(\Delta X_k) = \mathbb{E}[(\Delta X_k)^2] - \bigl(\mathbb{E}[\Delta X_k]\bigr)^2 = 1 \end{align} ...


2

Note that for the replicating portfolio to be self-financing it suffices that (1): $$\lambda_t=\frac{V_t-h_tS_t}{B_t}$$ where I have changed the notation by designating by $B_t$ the money market account: $$B_t=B_0e^{rt}$$ Hence, because the portfolio is self-financing, its dynamics are: $$\begin{align} dV_t&=\left(\frac{V_t-h_tS_t}{B_t}\right)dB_t+...


0

Showing that $(A_t^*)$ is a martingale does not really help you in understanding the distribution of $A_T^*$. Instead, the key is your previous question. Under $\mathbb{Q}\sim\mathbb{P}$, you have \begin{align*} S_t=S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right). \end{align*} Under $\mathbb{Q}_S\sim\mathbb{Q}$, you have \begin{align*} ...


5

Let $\mathbb{Q}$ be the risk-neutral probability measure which uses the risk-free bank account $(B_t)$ as numeraire. In general, $\mathrm{d}B_t=r_tB_t\mathrm{d}t$. In the Black-Scholes setting, $r_t\equiv r$, we have $B_t=e^{rt}$. The stock measure $\mathbb{Q}_S$ uses the compounded stock price $S_te^{qt}$ as numeraire and is defined via the Radon Nikodym ...


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