# Tag Info

11

By construction, the Itô integral, $I_t=\int_0^t X_s\text{d}W_s$, is a martingale if $\int_0^t \mathbb{E}[X_s^2]\text{d}s<\infty$. The martingale property, $\mathbb{E}_s[I_t]=I_s$ implies $\mathbb{E}[I_t]=I_0=0$. Because $W_s\overset{d}{=}\sqrt{s}Z$, where $Z\sim N(0,1)$, we indeed have \begin{align*} \int_0^t\mathbb{E}\left[\frac{1}{(1+W_s^2)^2}\right]\...

11

You need to rotate them so we can find some orthogonal axes. A simple way to think about this is by remembering that we can decompose the second of two brownian motions into a sum of the first brownian and an independent component, using the expression \begin{align} W_{t,2} = \rho_{12} W_{t,1} + \sqrt{1-\rho_{12}^2} \tilde{W}_{t,2} \end{align} where $\tilde{... 11 Besides @StackG's splendid answer, I would like to offer an answer that is based on the notion that the multivariate Brownian motion is of course multivariate normally distributed, and on its moment generating function. We know that $$\mathbb{E}\left(W_{i,t}W_{j,t}\right)=\rho_{i,j}t$$ i.e. an$N$-dimensional vector$X$of correlated Brownian motions has ... 9 I provide a solution in three steps. The first step carefully outlines how to split up the expectation and what new measures are used. This first step does not require any special model assumption and holds in a very general framework. I derive a formula for the option price that resembles the standard Black-Scholes formula. In a second step, I assume that ... 6 It is covered very nicely in Iain Clark's Foreign Exchange Option Pricing, A Practitioner’s Guide (pages 98-104). The book also contains references to the relevant literature including Feller's original paper. 6 The theory behind the actual reasoning is a bit complicated than the coverage in Hull's, but staying within the simple reasoning, the difference comes down to the following: The Brownian increments over the interval$dt$are normally distributed with mean zero and variance$dt$, so in terms of distribution, you can express the increments in terms of a ... 6 The call for the stock that can jump downward will be more valuable due to put-call parity. Suppose you have two stocks, both with a price of$100 and the same diffusive volatility. Stock A does not jump, whereas stock B can at some random time jump (for example) to zero. Clearly a put on stock B will be worth more, but the call must therefore also be worth ...

6

$$d Y \left(t\right) := d \left[\int_0^t{a \left(s\right)\mathrm{d}W_s}\right] = a \left(t\right) dW_t$$ Note that since $Y$ is a driftless process, it is a local martingale, and because $a$ is bounded, a true martingale. Its quadratic variation is given by $$\langle Y \left(\cdot\right)\rangle_t = \int_0^t{a^2 \left(s\right)\mathrm{d}s}$$ by definition ...

5

Integrating $W_t$ Consider the partition $t_i=it/n$ with $t_0=0$ and $t_n=t$. Then, by definition, \begin{align*} \int_0^t W_s\text{d}W_s &= \lim_{n\to\infty} \sum_{i=0}^{n-1} W_{t_i}\left(W_{t_{i+1}}-W_{t_i}\right). \end{align*} You can do the limit by using the identity $$W_{t_i}\left(W_{t_{i+1}}-W_{t_i}\right)=\frac{1}{2}\left(W_{t_{i+1}}^2-W_{t_i}^2-... 5 An alternative way is using the Stratonovich integral. By definition, we have$$\int_0^t X_s \, \circ dW_s = \lim_{n\rightarrow \infty} \sum_{i=1}^n \frac{X_{t_i} +X_{t_{i-1}}}{2}\left( W_{t_i} -W_{t_{i-1}}\right) \; \; (1)$$One can then show that for a deterministic smooth functions f and g we have:$$ \int_0^t g'(W_s)\, \circ dW_s = g(W_t)- g(W_0)\; \;...

5

For part 1 of your question, the short answer is no, calculating conditional density is a looong way of doing it. Possible but not the easiest. Here is the sketch for a shorter version. We note that $(X_{T/2},X_{T})$ is a jointly Gaussian vector with mean $\mu = (X_0 + aT/2,X_0 + aT)$, and the variance-covariance matrix $$\begin{pmatrix} b^2 T/2 & b^2 T/... 5 As shown in Credit Risk Modeling Notes (Bielecki, Jeanblanc, Rutkowski), Corollary 1.3.1, for t < s, we have:$$ P(\tau \leq s | {\cal F}_t) = N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right ) + {\rm e}^{-2\nu \sigma^{-2}Y_t} N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right ),$$where$$ Y_t = y_0+ \nu t +\sigma W_t, \: \sigma >0,...

4

It's just Girsanov's theorem. I suppose that under the risk neutral measure Q $$dS_{t}= r S_{t} dt + \sigma S_{t}dW_{t},$$ $$S_{t} = S_{0}\exp\left((r-\frac{\sigma^{2}}{2})T + \sigma W_{T}\right)$$ By multiplying by $e^{-rT}$ I have $e^{-rT}S_{T}$ which is a martingale so that I can change my measure under $Q$ to some equivalent probabilty $Q_{1}$ under ...

4

There's a lot left unspecified in this question, since it is stated without precision, but the effective idea of the answer given here is that those jumps introduce extra variation into the forward distribution of the underlying. And such variation is the bread-and-butter of option value. That said, the ambiguity in the question leaves room for other ...

4

Alternatively, we can use Ito isometry ($X$'s integrability and adaptability are assured by $a$'s boundness and adaptability, respectively): $$E[X_t|{\cal F}_s] = E[X_s\big|{\cal F}_s] + E\left[\left(\int_s^t a_udW_u\right)^2 - \int_s^t a_u^2du \big|{\cal F}_s \right]$$ $$= X_s + E\left[\left(\int_s^t a_udW_u\right)^2\big|{\cal F}_s\right] - E \left[ \... 4 With your SDE for F, I get:$$ dXdF = -a^2XFdt FdX = rFdt + aXFdW XdF = a^2XF dt -aXF dW$$So, adding up:$$ d(XF) = rF dt, $$giving$$ X_t = F^{-1}_t X_0 + rF^{-1}_t \int_0^t F_u du $$3 Our market has a tradeable asset S and a risk-less money market account B, that is, the numéraire of the risk-neutral measure. We assume the following standard conditions, which are widely applicable to most common models: We work in an Itô diffusion setting, and neglect jump modelling:$$\begin{align} & dS_t=\mu(t,S_t)dt+\sigma(t,S_t)dW^S_t \\ &...

3

Echoing some of the comments to the OP above, the only real difference between random walks and Brownian motions is a question of time frequency. IE a Brownian motion is just an aggregation of a (binary) random walk with higher frequency. Given both will always be at best an approximation of reality, asking for which is "better" becomes a bit of a ...

3

That the expectation is zero is often called the martingale property of Ito integral (see e.g. Oksendal Theorem 3.2.1). The formal proof consists of showing this for "simple" integrand functions and then generalising this by taking limits. This requires that the integrand process is adapted (i.e. not forward looking) and square integrable. Square ...

3

I believe your SDE has an unintended error. It should be: $$dr_t = a \cdot (b - r_t) \cdot dt + \sigma \cdot \sqrt{r_t} \cdot dB_t.$$ On the other hand, the Feller condition is discussed and explained in Section 10.2.1.2 (pg. 432) of the Andersen and Piterbarg book: Interest Rate Modeling. Hope it helps!

3

Write $X_t = W^2_t$, then you are trying to find $\int_0^t dX_t = X_t - X_0$. Now use Ito's Lemma to find the dynamics of $W^2_t$ and try to solve the SDE.

3

Let's take a standard Brownian motion $(B_t)$ and let's try to compute $\int_0^t B_s\mathrm{d}B_s$ in the Riemann-Stieltjes sense. Let $0=t_0<t_1<...<t_n=t$ be a partition and let $y_i=t_{i-1}$ or $y=t_i$ for $i=1,...,n$ be two intermediate partitions. Thus, \begin{align*} S^1_n(t) &= \sum_{i=1}^n B_{t_{i-1}}(B_{t_i}-B_{t_{i-1}}), \\ S^2_n(t) &...

3

Risk-neutral pricing is to help with relative value type questions: If I know the value of this what should the value of that be if it depends in some way on this. It doesn't help with absolute value type questions: Should I buy this or that, is the implied volatility too low or high etc. Those are generally "real world measure" type questions.

3

I think the question also brings up a common confusion with notation. I think it is incredibly unfortunate to use notation such as $dW(t)$ (unless it's part of a stochastic integral), and I get upset when I see it being used in textbooks. The definition of Brownian Motion is implicit and goes like this: (i) $W(t=0) = 0$ (ii) $W(t)$ is (almost surely) ...

3

Black scholes formula based on $S_t$ measure , theory, and formulas you mention are derived in detail in "Steven Shreve: Stochastic Calculus and Finance" draft pdf from 1997 , page 328 "stock price as numeraire".

3

Question 1 is answered in parts 1 through to 6: the idea is that each part slowly builds the tools required to derive the process equation for $S_t$ under the $S_t$ Numeraire. Question 2 & Question 3 are then answered in part 7. Part 1: Expectation of a function of a Random variable: Let $X(t)$ be some generic Random Variable with probability density ...

3

That looks correct, but a bit complicated. We know that under Black-Scholes with no dividends, $E^Q(S_t) = Forward = S_0 e^{rt}$ $e^{-rT}E^Q(\int_0^TS_tdt) = e^{-rT}\int_0^TE^Q(S_t)dt \\ = e^{-rT}\int_0^T S_0 e^{rt} dt = S_0 e^{-rT}\int_0^T e^{rt} dt \\ = S_0 e^{-rT} \frac{1}{r}(e^{rT} - 1) = S_0\frac{1-e^{-rT}}{r}$ It is straightforward to generalize to the ...

3

Providing only a sketch here, using Ito integral definition (and commuting limit, summations and expectation), the result boils down to studying the expectation term: $$E\left[ f_{t_{i-1}} (W_{t_i}-W_{t_{i-1}}) \cdot g_{t_{j-1}} (W_{t_j} - W_{t_{j-1}}) \right].$$ If the intervals don't overlap, $i\not= j$, and say $t_i \leq t_{j-1}$, then $f_{t_{i-1}} g_{... 3 (As said in the comments, you need to put down some of your thoughts regarding the question too, like specifying the tools/theorems you would use or actual attempts to apply them, even if you can only cover early steps, not just the question itself.) Hints: We are given$\$X_t^\theta:=\exp \left(\theta W_t-\frac{1}{2} \theta^{2} t\right)=\sum_{n=0}^{\infty} \...

2

That is because of sigma_tilde_squared == 0, You could add 0.01 at the add to avoid it == 0

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