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Hints: You know the vega of a digital call option formula: $V=-\frac{e^{-r(T-t)}}{\sigma} d_1 n\left(d_2\right)$ Where n is the standard normal density, which is positive. Sigma and exponential are also positive, so the sign of V is down to the sign of $d_1$. Which is negative when: $d_1 <0$ $\ln \frac{S}{X}+\left(r+0.5\sigma^2\right)(T-t)<0$ $S&... 6$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*} whereS_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$,... 6 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 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 ... 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. Essentially $$C = e^{-rT} \int_K^\infty \psi(S_T) dS_T$$ 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 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 A really simple and arbitrage free solution is to extrapolate flat volatility on the same moneyness. Let's say that you want an implied volatility for strike$K$at time$t<t_1$, and$t_1$is the first pillar on the surface. You look at the moneyness level$k=K/F_t$, then look for$K'$to get the volatility at the same moneyness level of the first pillar$...

4

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 ...

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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 $$V_{Binary}(K,\sigma) = -\... 4 risk-neutral. Really the forward measure. The price of the binary is struck at K is$$ Z P( F_T > K) $$with Z the discount factor and F_T the forward, and P the probability in the forward measure. If rates are deterministic, the forward measure and the risk-neutral measure will agree. 4 The price is, under the risk-neutral measure,$$ 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 ...

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An important practical reason is for hedging purposes. Consider a situation where the option is very close to maturity and the rate $R$ is fluctuating around the strike $K$, such that the option is alternating out-the-moneyness with in-the-moneyness. Compared to a standard call option, where the payoff is continuous in $R$ with a "smooth" ...

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Binary options are not traded on standard options exchanges and cleared OCC. Instead you are trading vs individual houses/web sites. These sites have been accused on not paying, freezing funds, and generally creating barriers to allow the customer to benefit from winning trades.

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Adding to the answer of @JoshK, many of the firms offering binary options often operate their scams by advertising their trading platforms on sites like Facebook. Their trading platform and website are designed to look very professional, making you think the company is legit. Some of the scammers go to the extent of creating YouTube accounts where they ...

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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 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 $$C(T,K) = C_{BS}(T,K,\Sigma(... 3 Ofcourse, It is always possible to find the implied volatility. The value of binary call is$$ {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 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^... 3 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 ... 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 There are a few extra things to consider here where you'll get a different answer if you ask a quant or a trader. If we have a european digital that pays \1 if the underlying is above 120 (S_0 = 100) at expiry, then yes i can hedge it with a call spread. This can be approximated with a call spread (with a notional of \frac{1}{\mathrm{d}K}, this was ... 3 Here is an intuitive explanation: you conclude that there is more chance of expiring otm than itm if there is skew but that isn’t correct. The atm volatility is unchanged vs the flat vol case, and atm is where you (you’re the stock) start from. Once you have moved from there (ATM), either you’re on the downside where vol is (only now) higher and the payoff ... 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 ... 3 The price close to 0.93 is correct, here is a reimplementation of both FD and analytic using QuantLib: import QuantLib as ql # World State for Vanilla Pricing spot = 50 vol = 0.2 rate = 0.01 dividend = 0.0 today = ql.Date(1, 9, 2020) day_count = ql.Actual365Fixed() calendar = ql.NullCalendar() # Set up the vol and risk-free curves volatility = ql.... 3 You can replicate the payout of a binary with a put spread with strike prices which are very close to one another. Higher skew makes the further out of the money put more expensive, which makes the put spread cheaper. 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 The value of an cash-or-nothing option is just the discounted expected payoff of the option. So the value of such a call should be$e^{-r (T - t)} N \mathbb{P} \left\{ S_T > K \right\}$, where$\mathbb{P} \left\{ S_T > K \right\} = \mathcal{N} \left( d_2 \right)$, and$N\$ is the cash agreed to be paid. The asset-or-nothing is a bit more complicated ...

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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("...

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