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1

First note that the price of binary call is related to the price of an ordinary call in any model by $$ BinC(T,K) = e^{-rT}\mathbb{E}^{\mathbb{Q}}[1_{S_T>K}] = - \frac{\partial}{\partial K}e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T-K)_+] = - \frac{\partial}{\partial K}C(T,K) $$ Now the volatility smile is implicitly defined by $$ C(T,K) = ...


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In the link you provided, by noting the construction of array p[], p0 and p1 are respectively the discounted $\texttt{down}$ and $\texttt{up}$ probabilities. Since $d=\frac{1}{u}$, then \begin{align*} p0 &= e^{-r \Delta T}\, \frac{u-e^{(r-q)\Delta T}}{u-d}\\ &= \frac{\big(u\,e^{-r \Delta T} -e^{-q\Delta T}\big)u }{u^2-1}, \end{align*} and ...


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It appears that you are plotting your analytical delta as a % of the delta of the underlying. This is why the delta converges to 100% As for the numerical delta, it could be that you are not adjusting for the DV01 of the underlying. This would explain why the numerical delta still increases as the option gets more in the money and why the distortion is ...


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The $P$ dynamics of the underlying asset are: \begin{align*} dS=S(\mu dt+\sigma dB_t) \end{align*} That has the following solution under the $\mathcal{Q}$ dynamics: \begin{align*} S_t=S_0 e^{(r-\frac{\sigma^2}{2})t+\sigma W_t} \end{align*} Where $W_t$ is the equivalent martingale with respect to the original geometric brownian motion. Define $Y_t=\int_0^t ...


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Yes it is known in closed form. See https://www.rocq.inria.fr/mathfi/Premia/free-version/doc/premia-doc/pdf_html/asian_doc/asian_doc.html section 5.1 which references an older Geman-Yor paper.


2

A few tips. First note that $e^{-rt}S_t$ is a martingale. So make it appear and then integrate by part to rewrite $\int S_u du$ as a stochastic integral. Finally use the Ito isometry property.


1

if we take a digital option and price under BS then you can do the whole thing by direct verification. i.e. $N(d_2)$ solves the PDE and converges to the final pay-off pointwise. So if the final pay-off has a finite number of jump discontinuities then subtract a linear combination of digitals to reduce to the continuous case.


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$E\{f(X_T)\}$ can still exist even if $f$ is not continuous. For example, $$ \begin{equation} f: x \mapsto \begin{cases} 1, \, x >= 3 \\ 0, \text{ otherwise} \end{cases} \end{equation} $$ Then $$ \begin{equation} E\{f(X_T)\} = P(\{X_T >= 3\}) \end{equation} $$ So, if you're example is a binary option which pays 1 if $X_T >= B$ and zero otherwise, ...


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It is because theta is not premium / days to expiration. Theta is a "local" decay, measure of current rate of option decay, which is not assumed to stay constant. In the example you provided, theta will be closer to zero (decay rate will slow down) as you approach expiration.


2

The price of the April option will be more than $5.00, correct. How much more depends on the implied volatility ($\sigma$) of the option and the interest rates ($r$). The higher $\sigma$ and $r$ are, the higher the time value of money and the value of the April option. I highly recommend playing around with this calculator to gain an intuitive ...


3

Simply put, no. Vega depends on a variety of factors (including the level/price of the underlying asset). However, vomma/volga/vega convexity (whatever you want to call dVega/dIV) is always positive. So as IV increases, the vega of an option increases - I think this might have been what you were getting at. It's important to understand that IV is an input ...


2

IV is one of the inputs for your option pricing model, vega measures the actual impact (e.g. in Dollars, Euros...) of any change in IV. Intuitively IV is the price of the option while vega is the sensitivity to IV. Bottom line: There is a clear distinction!


2

Because there are several non-linearities involved this depends very much on where you are concerning the level of volatility and time to expiry. But I think what you really want is to get some feel for the sensitivities involved, right? With the following demonstration you can play with all kinds of combinations of all parameters to get some intuition for ...


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By no arbitrage, $S(0)=(Su q+Sd (1-q))/(1+r)$ and $C(0)=((Su-k)^+ q+(Sd-k)^+(1-q))/(1+r)$. Simplifying and rearranging (and assuming $Su>k$), $$\left[\begin{array}{c} S(0) \\ C(0) \end{array} \right]=\frac{1}{1+r}\left[\begin{array}{cc} Su & Sd \\ (Su-k) & 0 \end{array} \right]\left[\begin{array}{c} q\\ 1-q \end{array} \right] $$ Clearly, ...


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Gatheral (Amazon) has a quite extensive discussion on that, and dives into calibration issues. In summary, what you describe appears to be less of a modeling issue, and more of a calibration problem. This is primarily because the model functions (such as the Heston model) are not by nature convex in their input parameters. This is simply result of the fact ...


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Your question concerns the term structure of volatilty. In this case, USO's vol term structure is inverted (downward sloping) since far-dated IV is less than current. Please keep in mind that the market determines the shape of the term structure and it can and will change over time. Currently, USO has seen a steep correction and there is some geopolitical ...



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