# Tag Info

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The text of your question doesn't actually match the question title. The answer to your title is of course yes binary options make sense. And as others have pointed out with binary options your reward is limited, and conversely the risk involved in writing them is less. To answer your additional question you can replicate a binary option with a tight call ...

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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} \... 5 The key difference besides the cap is that there is nothing in between: its 100 or nothing (binary!) - with traditional options you have S-K as long as S>K (for the call). You can find out more here: http://en.wikipedia.org/wiki/Binary_option 5 No. If you are long a vanilla option, your reward is unlimited. If you are short an option, your risk is unlimited. 5 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,... 4 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}}$$where N is the standard normal distribution function, F=Se^{(r-q)T} is the forward index price, S is the spot ... 4 No, there is an upper limit to a binary option's value, based on the interest rate and how much of the distribution can be packed under the payoff region. Essentially$$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 ... 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 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 They would make sense in certain narrow applications; one can perhaps think about scenarios where binary option might be the most efficient, quickest or easiest way to either benefit from a particular insight OR to hedge against some sort of event ... the real question is whether the volume in the markets for binary options will continue to sufficient to ... 2 For a long or short position in a vanilla put, the maximum risk/reward is known and capped. For a long position in a vanilla call, the maximum risk is also always known and capped. The maximum reward is therefore known and capped for a short vanilla call position. However, the reward is unbounded for a long vanilla call position, and therefore the risk is ... 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 Rather than thinking about the steps, think about the piecewise regions where your value is constant. When using the explicit scheme, time zero option value at any stock price for your simple digital option is basically just a function of which antecedent nodes (accounting for backwards timestepping) were above or below the strike. Slight modifications of ... 2 risk-neutral. Really the forward measure. The price of the binary is struck at K is$$ Z P( F_T > K) $$with Z the discount factor and F_T the forward, and P the probability in the forward measure. If rates are deterministic, the forward measure and the risk-neutral measure will agree. 2 First note that the price of binary call is related to the price of an ordinary call in any model by$$ 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(...

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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 ...

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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}^{*}\... 2 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 ... 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 ... 1 Two hints : The number of paths never going up to$3125$when starting from$1024$and stepping up by a multiplicative factor of$5/4$and down by a multiplicative factor$4/5$is the same as the number of paths starting from$0$and and stepping up by an additive factor$+1$and stepping down by an additive factor of$-1$and never going up to$5$Let$...

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the call version pays $$I_{S_T > K } S_T$$ the put version pays $$-I_{S_T < K } S_T$$ Subtract to get a pay-off $$S_T.$$ (ignoring the probability zero event of $S_T=K.$) So the prices subtract to give $S_0.$

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well, the current share price reflects fair value. So you'd expect it to be close to its expected price, but slightly below because of risk aversion and discounting. If it was very far off its expectation, it would either be over or under valued and people would trade accordingly.

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Sorry to disagree but if interest rates is 0, the binary is still not worth $1 now. Suppose spot$S(0) = 100$, assume$x = 110$and upon touch (whenever it happens as the option has no maturity) you receive one dollar. Suppose I buy 1 stock. If the barrier hits, i sell the stock and receive 110 USD. What if I buy N stocks at t=0? upon hit of barrier i ... 1 The skew plays an important part for pricing binaries. In S&P the VIX increases on declines and decreases on rises. We can explain a part of the premium by assuming the Black-Schole Call option captures the underlying volatility. Let us represent call option as$C(K,\sigma)$and binary$V_{Binary}(K,\sigma)$Then we can write$$V_{Binary}(K,\sigma) = \... 1 Black–Scholes usually assumes your time and volatility are annualised. Accordingly, when you calculate the volatility term you would usually annualise it to 252 or 260 (or however many trading days a year are applicable to your situation). Accordingly, the time remaining term of the Binary Option must also be expressed as a fraction of a year (again, 252, or ... 1 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 ...

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I believe this can be solved using the reflection theorem: prob (max S_t > x) = 2 * prob (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 prob (max P_t < y) = prob (max S_t < g(y) ) where the function g is the inverse of the function for ...

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