# Tag Info

7

Stochastics are usually applied in the field of derivatives pricing. In this setting the task is to price a derivative such that it fits into the landscape of tradable instruments (no-arbitrage). We work using the risk-neutral measure - usually denoted by $Q$. The measure is derived from other traded instruments. In risk analysis (e.g. calculate the VaR, ES ...

6

In the integral $$\int_0^t S_u dW^{*}_u \, ,$$ $dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$ is independent from the integrand $S_u$. So, $\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0$, since ...

6

It's a lemma! Ito's Lemma gives the change of coordinates rule for stochastic calculus. The multiplication rule is a shorthand way of expressing it.

5

The problem is equivalent to given to 2 independent standard normals $W$ and $Z$ the probability of $$W > 0, \text{ and } W+Z<0.$$ or $$W > 0, \text{ and } Z<-W.$$ Plotting this set we see it is the bottom half of the lower right quadrant. The probability of being in the lower right quadrant is clearly $0.25$ by symmetry. The probability ...

5

For any $s \geq t$, note that \begin{align*} r_s = r_t + \sigma\int_t^s dW_u + \int_t^s \theta_u du. \end{align*} Then, \begin{align*} \int_t^T r_s ds &= (T-t)r_t + \sigma\int_t^T\int_t^s dW_u ds + \int_t^T \int_t^s\theta_u du ds\\ &=(T-t)r_t + \sigma\int_t^T\int_u^T ds\, dW_u +\int_t^T\int_u^T\theta_u ds du\\ &=(T-t)r_t + \sigma\int_t^T ...

4

B1~N(0,1) and B2=B1+Z, for Z~N(0,1). From that E(B1*B1)=E(B1*B2)=1, E(B2*B2)=2. Therefore they are bivariate Gaussian with covariance matrix (1,1;1,2) therefore probability is around 12%, which is the volume over the bottom-right quadrant.

4

I thought this was an interesting example to add. It concerns a "ratio model" of habit (as opposed to a "difference" model of habit). See, for example, Abel (1990, American Economic Review). Let $$x_t = \lambda \int_{-\infty}^t e^{-\lambda(t-s)} c_s ds.$$ (For context, $x_t$ is a log habit index that is given by a geometric average of past consumption, ...

4

Maybe I'm missing something? Given $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$, you can write $f = (f_1,\ldots,f_m)$, where each $f_i:\mathbb{R}^n \rightarrow \mathbb{R}$. Apply Ito to each $f_i$ separately.

4

In this case it is just the notion that your payoff function should not explode at some point - made mathematically rigorous. Have a look at the following picture from wikipedia: Intuitively the Lipschitz condition (or Lipschitz continuity) ensures that your payoff function always remains entirely outside the white cone, so it cannot e.g. become ...

4

You know that $E\left[\int_{0}^{s}W_udu\right]=E\left[\int_{0}^{t}W_vdv\right]=0$. By definition \begin{align} & Cov\left(\int_{0}^{s}W_u\,du\,\,,\,\int_{0}^{t}W_v\,dv\right)=E\left[\int_{0}^{s}W_u\,du\int_{0}^{t}W_v\,dv\right]-0 \end{align} then \begin{align} & ...

4

Feynman–Kac Theorem: Assume that $F$ is a solution to the boundary value problem \begin{align} &F_t+\mu(t,x)F_x+\frac{1}{2}\sigma^2(t,x)F_{xx}-rF=0\\ &F(T,x)=\Phi(x), \end{align} Assume furthermore that the process $e^{-r_s}\sigma(s,X_s)F_s$ is in $\mathcal L^2$ where \begin{align} dX_s=\mu(s,x)ds+\sigma(s,x)dW_s, \end{align} then $F$ has the ...

4

Let $\{P_t \mid t \geq 0\}$ be a compound Poisson process, where \begin{align*} P_t = \sum_{i=1}^{N_t} (V_i -1), \end{align*} and $N_t$ is a Poisson process with intensity $\lambda$ and jump times $\tau_i$, $i = 1, \ldots, \infty$. Let $Y_i=\ln V_i$ and $f(x)$ be the density function. Then \begin{align*} P_t - \lambda t E(V_1) &= P_t - \lambda t ...

3

The logic from Bob Jansen is correct. The problem is abuse of ideas and notation the integral symbol from the deterministic world gets sloppily applied to random variables. Unlike normal $dt$, which is always positive, $dW_t$ can go 'backwards'. Thus increments of terms like $W_t dW_t$ have a first element that goes up and down with the second element ...

3

I am not sure if I understood your question correctly but I will try to answer it anyway. If you have a standard normal random vector $z \sim N(\mathbb{0},I_n)$ (where $z,0 \in \mathbb{R}^{n\times1}$ and $I_n \in \mathbb{R}^{n\times n}$ is the identity matrix) and you want to transform it into a multivariate normal $x \sim N(\mu,\Sigma)$ you do it the ...

3

What you need is to identify the distribution of the asset price $S_T$, conditional on the information set $\mathcal{F}_{t}$ at time $t$, for $0\leq t < T$. Note that \begin{align*} S_T &= S_t \exp\bigg(\int_{t}^T \Big(r_s-\frac{\sigma_s^2}{2}\Big)ds + \int_t^T\sigma_s dW_s \bigg). \end{align*} Let \begin{align*} P(t, T) = \exp\bigg(-\int_t^T r_s ds ...

3

Formally, this is a shorthand for the quadratic variation. For a more rudimentary definition, $\langle W, W\rangle$ is a process such that $W^2-\langle W, W\rangle$ is a martingale. Moreover, $\langle W, W\rangle_t$ is a limit, in probability, of the variation \begin{align*} \sum_{i=1}^n|W_{t_{i}}-W_{t_{i-1}}|^2, \end{align*} over the partition ...

3

Note that, for $0 \leq s < t$, \begin{align*} W_t^3 &= (W_t-W_s+W_s)^3\\ &= (W_t-W_s)^3 + 3(W_t-W_s)^2 W_s + 3 (W_t-W_s) W_s^2 + W_s^3. \end{align*} Moreover, \begin{align*} E\big( (W_t-W_s)^3 \mid \mathcal{F}_s\big) &= E\big( (W_t-W_s)^3\big)\\ &= 0,\\ E\big((W_t-W_s)^2 W_s \mid \mathcal{F}_s\big) &= W_s E\big( (W_t-W_s)^2\big)\\ ...

3

It is nearly a Bronwian motion. Just the variance is not correct: The question is more tricky than it seems. A Brownian motion has the distribution properties stated below, so does a linear combination of BMs. But after all it is a martingale in a certain filtration (set of information) which has to be defined. $B_t$ is a BM in its own filtration, so is ...

3

I think all they are doing is integrating and estimating $$P(|W_t| \leq 2) = \int_{-2}^{2} \frac{d}{dr} P(W_t \leq r) dr$$ so $$P(|W_t| \leq 2) \leq 4 \sup \limits_{r \in [-2,2]} \frac{d}{dr}P(W_t \leq r)$$ The normal density is maximal at zero and we are done.

2

$\sigma S$ is in units of dollars per square root of a unit of time. $\sigma$ is usually quoted as an annual or daily percentage. $dX ^2$ is in units of time, as $E[(dX)^2] = dt$. Here is an online tutorial which you may find helpful. EDIT by kotozna: $\sigma$ has dimensions 1/(square root of time) and $dX$ has dimensions square root of time. ...

2

Note that $X$ is a continuous martingale. Moreover, the quadratic variation is given by \begin{align*} \langle X_t, \, X_t\rangle = \int_0^t |\sigma_u|^2 du = c^2 t. \end{align*} That is, \begin{align*} \langle X_t/c, \, X_t/c\rangle = t. \end{align*} From Levy's characterization, $X/c$ is by law a Brownian motion, which we denote by $\beta$. Then, by law, ...

2

It appears that we need only to observe the following: \begin{align*} \lim_{\lambda\rightarrow 0}\frac{1}{\lambda}\int_0^{\lambda t}\sigma^2_u du &= \lim_{\lambda\rightarrow 0}\int_0^{ t}\sigma^2_{\lambda u} du\\ &= \int_0^{ t}\sigma^2_{0} du \\ &=\sigma^2_{0} t. \end{align*}

2

For the last question. We assume that \begin{align*} S_t = S_0 e^{(r-q-\frac{1}{2}\sigma^2)t + \sigma W_t}, \end{align*} where $W$ is a standard Brownian motion, $r$ is the interest rate, $q$ is the dividend yield, and $\sigma$ is the volatility. Then, \begin{align*} X_{u+a}-X_a &= (r-q-\frac{1}{2}\sigma^2)a + \sigma(W_{u+a}-W_u)\\ &\sim ...

2

The portfolio is self-financing. You simply forgot a term in $b$ and a $-t$ term in $V$: \begin{eqnarray} V_t &=& a_t S_t + b_t \beta_t = (2B_t ) (10+ B_t) + (- t - B_t^2 - 20B_t)1 \\ &=& 20B_t + 2B_t^2 - t - B_t^2 - 20B_t \\ &=& B_t^2 - t \end{eqnarray} Applying Ito's lemma \begin{eqnarray} dV_t &=& (2B_t dB_t + ...

2

I think to gain intution you have to understand that the same agents that value the stocks will value the options. And agents compensate for volatility by demanding higher expected returns. Therefore you should ask: Why are stocks priced as they are in the first place? In your example, the stock with higher volatility has much lower expected return. This ...

2

Yes and No. In the absence of arbitragers, the price of the option will be different for each speculator based on their drift expectations (and each speculator has a risk in his position and will limit his ability to trade large sizes to avoid bankruptcy) and the option price will converge to priced off a supply-and-demand driven drift expectation. ...

2

Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter. Also the delta-hedged call and the delta hedged put have to have the same value since they have the same pay-off. (Put-call parity) Yet any argument that the call should be worth more because of drift says that the put should be ...

2

if we forget about $S_0$, you are just trying to price a power option, i.e. an option on $S^\alpha$. By Ito $$d \log S^\alpha = \alpha d\log S = \alpha (r- q - \frac{1}{2}\sigma^2 ) dt + \alpha\sigma dW_t$$ This can be rewritten $$d \log S^\alpha = (r-q'-\frac{1}{2}\sigma'^2 ) dt + \sigma' dW_t$$ If you set $\sigma' = \alpha \sigma$ $q' = r ... 2 this is not the way to do it. The Black-_Scholes argument requires the underlying to be tradable.$S_{t}^{0.5}$is not tradable. Instead, recognize that the underlying is still$S_t$but the pay-off has changed to $$(\alpha S_{t}^{1/2} - \beta)_+$$ for appropriate constants$\alpha,\beta.\$ So the derivation of the BS equation still holds and the ...

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