Tag Info

New answers tagged

1

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 ...


0

Hope you will not mind if I place myself in continuous time. The discounted stock price at $T$ is $e^{-rT}S_T$. As you know that it is a martingale, you have that $\mathbf{E}^{\mathbf{P}}[e^{-rT}S_T | \mathscr{F}_t] = e^{-rt} S_t$ when $t\leq T$ which you can rewrite as $\mathbf{E}^{\mathbf{P}}\left[\frac{e^{-rT}S_T}{e^{-rt} S_t} | \mathscr{F}_t\right] = 1$ ...


0

In a one period model replication means that no matter which state of the model it holds that $$ a S_1+ b B_1 = D_1 $$ where $a,b$ are the quanties of stock and bond held and $S_1,B_1$ and $F_1$ are the prices of the stock, the bond and the derivative at $t=1$. In such a case the fair price of the derivative at time $0$ is $a S_0 + b B_0$ (not thinking about ...


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 ...


1

As you have guessed correctly, these type of questions can be answered using Ito's Lemma.We have: \begin{equation} d(M_t)= d(Z_t e^{\int_0^tF(Z_u)du})=d(Z_t) e^{\int_0^tF(Z_u)du}+Z_t d(e^{\int_0^tF(Z_u)du})+d(Z_t)d(e^{\int_0^tF(Z_u)du}) \end{equation} For the first two terms on R.H.S, we have: \begin{equation} d(Z_t) e^{\int_0^tF(Z_u)du} = (f(W_t)dW_t + ...


0

$F=0$ seems like a good choice.



Top 50 recent answers are included