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1

I'd like to give an alternative derivation not involving the clever (mystifying?) transformation to the heat equation and thus present a more general technique for solving constant coefficeint advection-diffusion PDEs. All we need is the Fourier transform: \begin{align*} \mathcal{F}[f] & = \int_{-\infty}^\infty e^{-i \omega y} f(y) dy, \end{align*} ...


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 BS model, with friction-free markets in continuous time, the cost of the hedging portfolio is the initial cost of setting up the portfolio. There are no costs over time as the hedging portfolio is self-financing: any purchase of the underlying is paid for by borrowing money and any selling of the underlying is invested at the risk-free rate. At ...


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 + ...


3

The above equation is the price of a call option. It has nothing stochastic inside it. It only depends on the current price and the time. So no Ito is needed. You should just compute the derivatives of your solution v (like you do for any deterministic multivariable function), plug them into the PDE and verify that it's satisfied.


3

I don't know the BS formula you are trying to use. The price is the expected value of the discounted payoff under the risk neutral probability measure (I.e. Under which S is a martingale) So the you need to compute the risk neutral probabilities for S to go up or down. The probabilities given in the problem have no impact. They are just there to trick the ...


1

Look at the B-S parameters for the dynamics of the stock. $\frac{dS}{S} = \mu dt + \sigma dt$ $\sigma$ is independent of strike in the B-S model, which means all derivatives priced assuming these dynamics should have the same volatility. This clearly is not the case given the existence of smile and skew. You can't assume the BS model produces the "fair" ...


0

The clue was to establish that there are typos in the script. Hence, I should have aimed to prove $\frac{Ke^{-r\tau}}{S^2\sigma\sqrt{\tau}}\Phi'(d_2)=\frac{\Phi'(d_1)}{S\sigma \sqrt{\tau}}$. Therefore, we have \begin{equation*} \frac{\partial^2 C}{\partial S^2}=Ke^{-r\tau}\mathbb{E}[\frac{\delta(S-U)}{U}] = Ke^{-r\tau} \int^{\infty}_0 \frac{\delta(S-u)}{u} ...


2

Since $Y=e^{(r-\frac{\sigma^2}{2})\tau + \sigma \sqrt{\tau}Z}$, then \begin{align*} xY > K \Leftrightarrow Z > -d_2, \end{align*} where \begin{align*} d_2 = \frac{\ln \frac{x}{K} + (r-\frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}. \end{align*} Consequently, \begin{align*} e^{-r\tau}\mathbb{E}\big(Y \mathbb{1}_{\{xY >K\}} \big) &= ...


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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