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The price of a binary option, ignoring interest rates, is basically the same as the CDF $\phi(S)$ (or $1-\phi(S)$ ) of the terminal probability distribution. Generally that terminal distribution will be lognormal from the Black-Scholes model, or close to it. Option price is $$C = e^{-rT} \int_K^\infty \psi(S_T) dS_T$$ for calls and P = e^{-rT} \... 6 I_{\{S_{T}-K>0\}} is NOT independent of \mathcal{F}_{t}, since \begin{align*} S_T=S_t \, e^{(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^*-W_t^*)}, \end{align*} where S_t \in \mathcal{F}_t, though e^{(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^*-W_t^*)} is independent of \mathcal{F}_t. However, since W_T^*-W_t^* is independent of \mathcal{F}_t,... 5 No, there is an upper limit to a binary option's value, based on the interest rate and how much of the distribution can be packed under the payoff region. EssentiallyC = e^{-rT} \int_K^\infty \psi(S_T) dS_T$$for calls and$$ P = e^{-rT} \int_0^K \psi(S_T) dS_T$$for puts. Neither of the integrals can ever exceed 1.0 and often they take on a ... 5 Are you sure you are using the correct pricing formula. For a binary (digital) call that pays 1, the simple Black-Scholes price at time t=0 is$$ C_d = e^{-rT}N(d_2)d_2 = \frac{\text{ln}(F/K) - \frac1{2}\sigma^2T}{\sigma \sqrt{T}}$$where N is the standard normal distribution function, F=Se^{(r-q)T} is the forward index price, S is the spot ... 5 Let f_0(S_T) =f(S_T|S_0) be the risk-neutral PDF for the underlying asset price at time T (conditional on the price S_0 at present time t=0). The probability that the price is above a strike price K at time T is$$P(S_T \geqslant K) = \int_K^\infty f_0(x) \, dx.$$This is just definitional regardless of the shape of the distribution (eg. ... 4 You can also infer the value of your binary option from the value of a European call option with the same strike and time-to-maturity by going long on a call with strike K and time-to-maturity \tau, and short on a call with strike K+\Delta K and the same time-to-maturity. If you hold \frac{1}{\Delta K} of that portfolio, then as \Delta K goes to ... 4 The price is, under the risk-neutral measure,$$ P_t = e^{-r(T-t)}\mathbb E[S_T^1 \mathbb 1(S_T^2\le K)\mid \mathcal F_t].$$Since the risk-neutral asset processes are independent geometric brownian motions, S_T^1 and S_T^2 are conditionally independent given \mathcal F_t. So the conditional expectation factors and you get$$ P_t = e^{-r(T-t)}\mathbb ...

3

I think that I found correct answer to my question. We have the following theorem: Theorem. If $X$ is independent of $\mathcal{G}$, $Y$ is $\mathcal{G}$ - measurable and $\phi(x,y)$ is bounded function then: $$E\left[\phi(X,Y)|\mathcal{G}\right]=E\left[\phi(X,y)\right]$$ In my problem I want to calculate the following formula: $C_{t}=\mathbb{E}^{*}\... 3 The skew plays an important part for pricing binaries. In S&P the VIX increases on declines and decreases on rises. We can explain a part of the premium by assuming the Black-Schole Call option captures the underlying volatility. Let us represent call option as$C(K,\sigma)$and binary$V_{Binary}(K,\sigma)$Then we can write $$V_{Binary}(K,\sigma) = -\... 3 Ofcourse, It is always possible to find the implied volatility. The value of binary call is$$ {e}^{-r(T-t)}N(d_2) $$where$$ d_2=\frac{ln(\frac{S}{E})+(r-D-\frac{\sigma^2}{2})\tau}{\sigma\sqrt\tau} $$Now, there is nothing that can ever ever stop the newton raphson method to find a \sigma for which the value of binary call is given and is positive ... 3 risk-neutral. Really the forward measure. The price of the binary is struck at K is$$ Z P( F_T > K) $$with Z the discount factor and F_T the forward, and P the probability in the forward measure. If rates are deterministic, the forward measure and the risk-neutral measure will agree. 3 First note that the price of binary call is related to the price of an ordinary call in any model by$$ 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) = C_{BS}(T,K,\Sigma(... 3 I have a mathematical proof with no graphs or pictures. Suppose$r=0$, what we want is to see what happens if volatility changes in$E^Q[1_{S_T>K}]$. The latter quantity is$Q(S_T>K)=Q(\log S_T > \log K)$. Under Q, we know that$S_T=S_0 \exp\left(-\frac12 \sigma^2T + \sigma W_T\right)$, so$\log S_T$is distributed as$ N(\log S_0 -\frac12\sigma^...

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all of the volatility effects on a binary option struck at 105 with a one dollar payoff are approximately the same as the volatility effects on the following portfolio of options: short 100 of the 104.99 calls / long 200 of the 105 calls / short 100 of the 105.01 calls

3

Disclaimer: This answer derives the prices of two different binary options within the Black/Scholes framework. Note that this is not an appropriate valuation model to use for non-European contracts in most real-world markets. Up-and-In Binary Call After reading your question for a second time, I agree with Quantuple's comment that you seem to be looking ...

3

If we are talking about brokers who making markets for https://en.wikipedia.org/wiki/Binary_option than I would guess that they aren't hedging at all. It's very common that maturities are in a timeframe of seconds or minutes. In my opinion returns are completely random in those timeframes.

3

$S_T$ is log-normal distributed and therefore skewed. In particular $E[S_T]=S=110$ (no drift), but $Q(S_T>S)<Q(S_T<S)$. For example if S=K=100 you don't get a value of 0.5 as you might expect, but a lower value since you have norm.cdf($-0.5\sigma\sqrt{T}$)$<$ norm.cdf($0$)=0.5.

3

As Daneel mentioned in his comment, you can't simply split your expectation of product into a product of two expecations as the two quantities are far from being independent... Now, to answer your question w.r.t. how you could compute the expectation of the joint event of being in the money while having hit the barrier, you were right in using the reflexion ...

2

5 minutes is a very short time period! If you have access to real time data of Implied Volatility and transaction Volume of the underlying of your option than you can take a look to the following article: Volatility Forecasts, Trading Volume, and the ARCH versus Option-Implied Volatility Trade-off In this article, the authors use the information from ...

2

First of all, don't forget that there are two different probability measures at play here: the frequentist market measure that reflects actual observation of the market "in the long run" and the market-neutral martingale measure which is pertinent for pricing options. More or less, we can take the frequentist measure, and "back out" the effects of market ...

2

I believe this can be solved using the reflection theorem: $$P(\max S_t > x) = 2 P (S_T > x)$$ Hence the required densities can be obtained solely from the distribution of $S_T$. There is a one to one correspondence between $\max P_t$ and $\max S_t$, so that $$P (\max P_t < y) = P (\max S_t < g(y) )$$ where the function $g$ is the inverse of ...

2

You can derive these formulae by tweaking the black scholes derivation. If you are using PDE method, you will use different boundary conditions. If you are using integration over the risk neutral probability , you will use a different payoff function but the same risk neutral density. Alternatively , you can observe that these payoffs are combinations ...

2

It is all in the code:: Rcpp::List rl = Rcpp::List::create(Rcpp::Named("value") = opt.NPV(), Rcpp::Named("delta") = opt.delta(), Rcpp::Named("gamma") = opt.gamma(), Rcpp::Named("vega") = (excType=="european") ? opt.vega() : R_NaN, Rcpp::Named("...

2

As I mentioned above, I am not sure what the variable $r$ is. If we ignore that, or assume the questioner wanted to say its the risk free interest rate, then it has no effect on the number of paths. Then it is clear that after 50 steps going from \$1024 to \$2500 requires a net of 4 up movements with the given $x=y^{-1}=1.25$. Thus the number of steps ...

2

The main difference is that with a binary option you are betting on a real economic risk, that exists independently of the bet. (For example, even if options did not exist, if stock prices go down pension funds will suffer losses, if the price of oil goes up Venezuela and Saudi Arabia will benefit). With gambling you are betting on an artificial risk, such ...

2

Your question is essentially the same as this. The approximation $$V_{t + 1} \approx V_t + \Delta_t \left( S_{t + 1} - S_t \right)$$ is only accurate when $S_{t + 1} - S_t$ and $\Delta t$ are small. See also my answer to this question. It provides some references that show that the delta is bounded by the slopes of the payoff ...

2

The goal of this exercise is to replicate the payoff of the Secured Barrier Call by a linear combination of the known products: European up-out call (cost 12), digital strike 33 (cost 0.73) and digital strike 50 (cost 0.7). Looks to me it is sufficient to buy: 1x up-out call 50 x digital strike 50 The payout at expiry of this linear combination would be: ...

2

My actual area of research is equity securities, however, I was once called upon to evaluate an algorithm for fx trading as part of a due diligence. FX isn't centrally traded. There isn't a single feed. They cannot match because there isn't a unified price. It will gum up your regression. It is probably impossible to find out where they are getting ...

2

The price is $e^{-r(T-t)} \mathbb{P}(S_{T}^{1} > S_{T}^2) =e^{-r(T-t)} \mathbb{P}(S_{T}^{1} / S_{T}^2 >1)$ The crucial point is that the ratio of two log-normals is log-normal even when they are not perfectly correlated so it just comes down to a cumulative normal. We assume vols are $\sigma_1$ and $\sigma_2$. Correlation between driving BMs is $\... 2 As you say, you simply differentiate with respect to$K$. Assuming your binary's maturity is$T$, note that in a Black-Scholes framework with constant risk-free rate$r$, by the Breeden-Litzenberger equations:$\$ \begin{align} \text{Binary}&=\lim_{\epsilon \rightarrow 0}\frac{-C(K+\epsilon)+C(K)}{\epsilon} \\[6pt] &=-\frac{\partial C}{\partial K}(K) ...

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