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7

Options on interest rates futures in the listed markets are always traded 1-yield (100-yield) just like the futures which are traded 1-yield. So negative rates aren't an issue and its always black volatility. In the OTC market, both normal and black volatility are quoted, but the common practice is to use black volatility is what is way more frequently used....


6

Here couple pointers to push you back on the right path (so I hope): Start with the payoff function and hence $S(T)$, which consists of $(W(T)-W(t))$ , $W$ being a Brownian Motion under the risk neutral measure) you can greatly simplify by working with a standard normal random variable: $$Y = \frac{-(W(T)-W(t))}{\sqrt{T-t}}$$, which helps to get rid of ...


6

You ask 2 questions and I try to answer: 1) Why do we use geometric Brownian motion ($\ln S_t-\ln S_0$ is normally distributed)? In this case you have $$ S_t = S_0 \exp( (\mu-\sigma^2/2) t + \sigma B_t), $$ which means that you model positive prices. Furthermore the log-return $$ \ln(S_t/S_0) = (\mu-\sigma^2/2) t + \sigma B_t, $$ is normally distributed. ...


6

please go to {drvd} BVOL Equity Implied Volatilities Calculations paper. Disclamer: I was working for Bloomberg, that is as far we disclosed.


6

Quantiles are preserved under monotonic transformations, hence the quantile for $Y$ is simply the exponential of the quantile of $X$, no need for corrections whatsoever (see here for instance). Put otherwise, let $q$ denote the quantile $\alpha$ of $X$ i.e. $$\Bbb{P}(X \leq q) = \alpha$$ then \begin{align} \Bbb{P}( X \leq q ) &= \Bbb{P}( \underbrace{\...


5

The Black-Scholes-Merton (1973) model implies that the prices of the underlying asset at maturity $S_T$ are log-normally distributed $$ln(S_T)\sim N\big[ln(S_0)+(\mu-\frac{\sigma^2}{2})T,\;\sigma^2T\big]$$ so that the logarithmic returns to maturity $ln(\frac{S_T}{S_0})$ are normally distributed $$ln(\frac{S_T}{S_0})\sim N\big[(\mu-\frac{\sigma^2}{2})T, \;\...


4

The smile is there exactly because the model is wrong. The reason it's used though (despite being wrong) is that it provides a convenient space to look at the underlying - the vol* The (undiscounted) value of an option is given by: $$ \int_0^\infty \mathrm{PDF}(s) (s-K)^+ \mathrm{d}s $$ where $\mathrm{PDF}(x)$ is the real probability distribution of the ...


3

The implied Black-Scholes skew will be downward sloping in the limit on both the left and the right. (I believe @Gordon's derivation claiming upward slope may have a sign error somewhere). Left Side For the left side it is sufficient to note that the lognormal model has no density below zero while the normal model has strictly positive density in that ...


3

Since $S_T = S_0 + \sigma W_T$, \begin{align*} C &:= E\left((S_T-K)^+ \right)\\ &= E\left((S_0+\sigma W_T-K)^+ \right)\\ &=\int_{\frac{K-S_0}{\sigma \sqrt{T}}}^{\infty}(S_0+\sigma\sqrt{T} x-K) \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=(S_0-K)\Phi\left(\frac{S_0-K}{\sigma \sqrt{T}}\right)+\frac{\sigma\sqrt{T}}{\sqrt{2\pi}}e^{-\frac{(S_0-K)...


3

Note that \begin{align*} E(K) &= E\big(\exp(\ln K) \big)\\ &=\exp\Big(E(\ln K) + \frac{1}{2}\sigma_k^2 \Big),\\ E(L) &= E\big(\exp(\ln L) \big)\\ &=\exp\Big(E(\ln L) + \frac{1}{2}\sigma_l^2 \Big),\\ E\Big(\frac{1}{P}\Big) &= E\big(\exp(-\ln P) \big)\\ &=\exp\Big(-E(\ln P) + \frac{1}{2}\sigma_p^2 \Big), \end{align*} and \begin{align*} ...


3

Generally Bloomberg is very open with their methodologies. Look up the documentation as recommended above, and if you have further questions you can ask HELP HELP to put you in touch with someone on their quant development team for more details. As long as you are a paying subscriber it should be no problem.


3

Interest rate options (swaptions, caps, floors, spread options, mid-curves, etc) that are traded over-the-counter (OTC), as well as those listed on the Liffe/CME exchanges, have been quoted using Normal volatility (basis points, annualised) for quite some time for several reasons, not least of which is the lack of a real zero-bound in yields that you ...


3

To compute the variance $$\text{Var}\left(\int _0^T e^{W_t} dt \right),$$ we need to compute \begin{align*} E\left( \left(\int _0^T e^{W_t} dt \right)^2 \right) &= \int_0^T\!\!\!\!\int_0^T E\left(e^{W_s+W_t} \right) ds\,dt. \end{align*} Note that, for $0 \le s, t \le T$, \begin{align*} W_s+W_t = \begin{cases} W_t -W_s + 2 W_s, & \text{ if } s \le t,\...


3

As @Rustam notes, "correlation" of deterministic functions in the sense you describe is a special case of allowing $\mu$ and $\sigma$ to have a term structure of arbitrary shape. Since the latter is easy to treat, no one bothers with restricted forms of it. Now, there quite a few people who deal with models that let $\sigma$ change with $S$. I am thinking ...


3

What you have to start with is: $$dS_t=\mu S_t dt + \sigma S_t dW_t$$ where $W_t$ is a standard brownian motion (SBM). You want to solve for $S_t$, so how would you proceed? If you integrate both sides of the equation between 0 and $T$, you get: $$S_T - S_0= \mu \int_0^T S_t dt + \sigma \int_0^T S_t dW_t$$ Okay and then what? The fact that you have $...


3

The most likely reason I can think of is the ease of computation. Gerald Appel developed the MACD in the late 1970's, when computing resources were very limited. When doing calculations by hand, on paper, it's much easier to take the difference of two simple (or exponential) moving averages than the log of their quotients.


3

BS assumes prices NOT returns are log-normally distributed. Why making that assumption? 1.log-normal is not perfect but OK to fit potential prices distribution. 2.The nature of log-normal distribution will force the left tail to be above zero. 3. There are definitely distributions work better than log-normal in terms of fitting stock price data, but that ...


3

In the Black Scholes (1973) model, the stock price is assumed to follow a geometric Brownian motion $\mathrm{d}S_t=S_t\mu \mathrm{d}t + S_t \sigma \mathrm{d}W_t$. If you solve the SDE, $(S_t)$ is log-normally distributed for every $t$. Alternative, you can model the returns by a normal distribution and then take the exponential function to obtain the stock ...


2

Actually it is quite simple to demonstrate Ito's correction term in a binomial tree. Details can be found in my new paper (p. 8-10): von Jouanne-Diedrich, Holger: Ito, Stratonovich and Friends (April 21, 2017) Abstract This exposition should provide you with the bigger picture of stochastic calculus, especially stochastic integrals. It heuristically ...


2

One way to start thinking about this is to work out a couple of Discrete versions of Ito's lemma Øksendal (6th edition) Example 3.1.9: almost surely, $$ B_t^2 - t = \int_0^t 2B_s dB_s $$ This has a discrete version which holds everywhere: let $X_n=\pm 1$ and $S_n=\sum_{i=1}^n X_i$, then $$ S^2_n-n = 2\sum_{i=0}^{n-1} S_i X_{i+1} $$ To verify ...


2

I doubt you can do this. Correction term appears in Ito because Brownian motion has infinite variation (non zero quadratic variation). In discrete and therefore finite models you cannot observe this phenomenon.


2

There are many ways answering this, here is one: We assume the asset price at $t=T$, $S_T = S_{T-1} \times (S_T / S_{T-1})$. Assuming continuous compounding, we can write, $S_T = S_{T-1} \times \exp(R_{T-1})$. Working the same way for the previous period, we get $S_{T} = S_{T-2} \times \exp(R_{T-1}+R_T)$. Working all the way back to the initial value of ...


2

if they are stocks, this problem is called pricing a Margrabe option and it is generally solved by change of numeraire. Take $S_2$ to be the numeraire. Then the value of the option is $$ S_2(0) \mathbb{E}_{S_2}( (S_1(T)/S_2(T)-1)_+) $$ where the expectation is taken in the measure that has $S_1/S_2$ as a martingale. Since it's a martingale and log-normal at ...


2

you got a typo. It should be 40.886 in your last equation. Then $\sigma$ should match. Also, If $\alpha$ means annualized log return, it should be $\mu\,t = \alpha - \frac 1 2\sigma^2\,t$ So in your last two equations, the first term should be $\frac \alpha {250} - \frac 1 2 \sigma^2$


2

Yes, your steps are valid This is a wrong use of the term "quantile". Here you need to compute a probability (through the normal cdf) and not a quantile (i.e. the value of a random variable corresponding to a given level of the cdf, e.g. the quantile 0.5 (or percentile 50%) is the median)


2

If $S_t$ is stochastic process and follow geometric Brownian motion with following SDE: $$dS_t=\mu S_t dt + \sigma S_t dW_t$$ then $S_T$ follows lognormal distribution, such that: $$S_T|S_t \sim logN\left(lnS_t+ (\mu - \frac{\sigma^2}{2})(T-t), \quad \sigma^2(T-t)\right)$$ or $$lnS_T|S_t \sim N\left(lnS_t+ (\mu - \frac{\sigma^2}{2})(T-t), \quad \sigma^2(T-...


2

The random variable $$X = \sum_{k=1}^{n} A_k\exp(\mathcal{N}(t_k\mu-\sigma\sqrt{t_k}/2,\sigma)))$$ emerges as a weighted sum of individual random variables that are log-normally distributed. Unfortunately, even if we assume that the individual r.v. involved in the sum are independent, a sum of log-normals (here $X$) possesses no analytically tractable ...


2

To my knowledge, adding a minus 1 does not transform a log normal variable into an normal distributed variable. The only thing that I can think of which make sense is the log normal represents a price ratio $\frac{P_t}{P_0}$ (for instance if the price process is a geometric browninan motion). In this case adding -1 does transform price ratio into arithmetic ...


2

If I'm understanding you correctly, the log returns are normal, but the simple returns are not. While I'm surprised your plots are that different, simple returns will not be normal even if the log returns are normal; they will instead be (shifted) log-normal. If $S_t =S_0 e^{y\sqrt{t}}$ where $y$ is Gaussian, then the simple return is $\frac{S_{t+1}-S_{t}}{...


2

Under the risk-neutral probability measure $\mathbb{Q}$, the logarithmic return is normally distributed with \begin{equation} \ln \left( \frac{S_T}{S_0} \right) \sim \mathcal{N} \left( \left( r - \frac{1}{2} \sigma^2 \right) T, \sigma^2 T \right). \end{equation} Thus, \begin{eqnarray} V_0 & = & \frac{1}{T} e^{-r T} \mathbb{E}_\mathbb{Q} \left[ \ln ...


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