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0

By using the fact that the brownian integral has expected value $0$, we find \begin{eqnarray*} & & \left(1-e^{-\rho dt}\right)v(X_{t}) \\ &=& E_{t}\left\{ \int_{t}^{t+dt}e^{-\rho(s-t)}u(X_{s})ds+e^{-\rho dt}\left[\int_{t}^{t+dt}v'(X_{s})dX_{s}+\frac{1}{2}\int_{t}^{t+dt}\sigma(X_{s},s)^2 v''(X_{s})ds\right]\right\} \\ &= & E_{t}\left\{ ...


1

We have $$ V_t = a_t S_t + b_t \beta_t. $$ By Ito's product rule, \begin{align*} dV_t & = d(a_t S_t) + d(b_t \beta_t) \\ & = a_t dS_t + S_t da_t + da_t dS_t + b_t d\beta_t + \beta_t db_t + db_td\beta_t. \end{align*} Since $da_t$ and $db_t$ have no $dW_t$ term, the cross terms are both zero and we have \begin{align*} dV_t & = a_t dS_t + S_t da_t ...


1

If $\sigma=0$ there is no randomness: the spot follows a single deterministic path. That is, the measure consists of a point mass at that path. Any equivalent measure can again only give a point mass at that same path, with the same drift. So in this case we must have $\mu = r$ to have an equivalent martingale measure. This is arbitrage free, but there ...


1

In the Black-Scholes model, you would have $d S_t = \mu\, d t + \sigma\, d W_t$ where $W$ is a Brownian motion. So if $V_t = a_t S_t + b_t \beta_t$, then $$ dV_t = a_t\, d S_t + S_t\, d a_t + da_t\,dS_t + b_t\,d\beta_t + \beta_t\,d b_t + db_t\, d\beta_t $$ by the product rule. In your case, when $a_t = 1-t$ you will have $$ dV_t = (1-t) \, dS_t - S_t\, dt + ...


1

This is a special case of the question of why $$ \int_0^T f(t) dW_t $$ is normally distributed for a continuous function $f(t).$ This Ito integral can be approximated by a sum $$ \sum_{i=0}^{N-1} f(i T/N) (W_{(i+1)T/N} - W_{i T/N}) .$$ The Brownian increments $(W_{(i+1)T/N} - W_{i T/N})$ are independent normally distributed random variables. The key point ...


1

For simplicity, We assume that $\alpha$ is a positive constant. You need to show that, for any $t>0$, \begin{align*} \int_0^t e^{\alpha u} dW_u \end{align*} is normally distributed. Consider the process $\{X_t, t \geq 0\}$, where \begin{align*} X_t = \frac{1}{\sqrt{\frac{1}{t}\int_0^t e^{2\alpha u} du}}\int_0^t e^{\alpha u} dW_u, \end{align*} for ...


4

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 ...


1

Let $Z_1,Z_2\sim N(0,1), B_1=Z_1,B_2=Z_1+Z_2.$ Construct a random variable $Y$ as following: $$\left\{ \begin{array}{cc} Y=1 & B_1<0, B_2>0\\ Y=0 & otherwise \end{array} \right. $$ Note that $\mathbb{P}(Z_1+Z_2>0\mid Z_1<0)=\mathbb{P}(Z_2>-Z_1\mid Z_1<0)=\mathbb{E}Y$. Use that to construct the integral. Everything that is not ...


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.


0

It might be easier to go the other way: start with $$ d\mathcal{E}_t = \mathcal{E}_t dX_t $$ apply Ito to the $\log$ function $$ d\log(\mathcal{E})_t = \frac{1}{\mathcal{E}_t}d\mathcal{E}_t - \frac{1}{2} \frac{1}{\mathcal{E}_t^2}d\langle\mathcal{E},\mathcal{E}\rangle_t = \frac{1}{\mathcal{E}_t}\mathcal{E}_tdX_t - \frac{1}{2} ...


0

For question I, the identity \begin{align*} \rho_t = \exp\big(-\lambda_t W_t - \frac{1}{2} \lambda_t^2t\big) \end{align*} does not appear correct, unless $\lambda_t$ is a constant. For question II, yes. If $X_t = -\int_0^t \lambda_s dW_s$, then $\langle X \rangle_t = \int_0^t \lambda_s^2 ds$. For question III, you need to note that \begin{align*} \langle X ...


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, ...


0

So I have a "+" sign for the second term (not negative) dr =r[(θ(t)+ d(lnσ)/dt * lnr + 1/2*σ^2)dt + σdW] I left out the subscript t's..... You can let V = log r then Apply Ito and solve for A and B... where B = r*σ


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 + ...


0

Your choice of $a_t$ and $b_t$ is feasible. For a self-financing portfolio, the units invested should be static within an infinitesimal time interval, that is, no extra investing or withdrawing during this period. In other words, the portfolio value changes only through its underlyings. For a further discussion, See the article ...


0

For simplicity, we assume the necessary positivity, and then we can ignore the absolute signs. Note that \begin{align*} \big(C_1 - a e^{-bt} \big) d\omega = \big(C_1 - a e^{-bt} \big) f(t) dt + cbe^{-bt} \omega dt. \end{align*} That is, \begin{align*} \big(C_1 - a e^{-bt} \big)\,d\omega - cbe^{-bt} \omega \, dt= \big(C_1 - a e^{-bt} \big) f(t)dt. ...


0

User9403 nails it! Intuitively higher expected rate of return => higher stock price => higher (call) option price!


-2

"Why the expected return rate of a stock has nothing to do with its option price?" It has everything to do with the option price! The option price is a function of the stock price. If the expected rate of return on the stock price declines, the stock price will decline as will the option price. "Suppose I have two stocks A and B, the price is the same ...


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 ...


1

Practically, it is very difficult to get a measurement of a stock's true drift while there are very well-documented processes to estimate volatility. It is therefore very convenient mathematically to select the risk neutral pricing measure that eliminates idiosyncratic drift. At its heart, Black Scholes constructs a dynamic, replicating portfolio for an ...



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