28
You can directly imply a probability distribution from a volatility skew.
Note that, for any terminal probability distribution $p(S)$ at tenor $T$, we have the model-free formula for the call price $C(K)$ as a function of strike $K$
\begin{equation}
C=e^{-rT} \int_0^\infty (S-K)^+ p(S) dS
\end{equation}
Therefore we can write
\begin{equation}
e^{rT} \...
12
There are a number of different tests that are generally used to compare samples to different distributions, such as Jarque-Bera, Anderson-Darling, and Kolmogorov–Smirnov (see this related question).
In your case, with just the standard deviation and mean, there isn't a whole lot to say. You need to assume a distribution (e.g. normal). You would be able ...
12
Interest rates in general are far from independent and identically distributed. A high interest rate observation is quite likely to be followed by another high observation, and the volatility is likely to be higher as well. Interest rates are also mean reverting, as in most real-world situations (at least for developed markets) interest rates rarely rise ...
10
Upon close reading, this appears to be 3 (interesting) questions, not one. I'm not sure if the mods have the tools needed to split it up, so I'm just going to write down the three questions as I see them and then deal with them one by one. Note, it is simpler for me to talk about variance instead of volatility. This has no material impact on the answer.
...
7
What you refer to as the 99.5th percentile is known as the "Value-at-Risk." You are correct that you will need to make a distributional assumption, and there is a popular and well-researched approach to this problem, though I'm not certain it could be called "standard." I would recommend you use the "truncated Levy flight" distribution. James Xiong at ...
7
Well there are two main things to consider here.
Many implementation of Black-Litterman use the market portfolio and the ex post volatility and correlation structure to back out implied returns to use as prior. As far as I know, there is no standard way to reverse-engineer the optimization problem in the presence of nonnormal markets. (the first guess is ...
6
You could try using the Gaussian Affine Term Structure Models (GATSM), with the right boundary conditions to stop rates being negative (in the style of their Black implementation). See, for example, Monika Piazzesi, the "Affine Term Structure Models" if you want to enter/modify the basis or the work of Krippner, for example "Measuring the stance of monetary ...
6
Be careful, remember that the mean and the standard deviation don't tell you the whole story: http://en.wikipedia.org/wiki/Anscombe%27s_quartet
6
The best introduction to copulas I know, i.e. with rigour and intuition, is the following.
THE QUANT CLASSROOM BY ATTILIO MEUCCI
A Short, Comprehensive, Practical Guide to Copulas
Visually introducing a powerful risk management tool to generalize and stress-test correlations
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
I think an extremely interesting strand of research on this topic is represented by extensions of vine copulas with time-varying parameters.
For vine copulas in general have a look at this site from the Technische Universität München:
Vine Copula Models
One of their research projects, which is the most relevant in this context, is:Time varying vine copula ...
5
I found Coping With Copulas by Thorsten Schmidt really helped me to get a more basic understanding of copulas. As well as looking at some simple examples in R and thinking about different directions the transformations can happen.
To answer your actual question I'll attempt to describe the steps involved as simply as I can.
Let's say you use the copula ...
5
In the theory of copulas you want to model a multivariate (often bivariate) distribution and keep the marginals fixed.
Thus you have random variables $X$ and $Y$ with cdf $F_X(x) = P[X \le x]$ and $F_Y(y) = P[Y\le y]$ and you want to find some $F_{X,Y}(x,y) = P[X \le x, Y\le y]$ such that when you look at marginals you get $F_{X,Y}(x,\infty) = F_X(x)$ and ...
4
Exponential distribution, although it's a good distribution for modeling non-negative numbers, doesn't make sense here since it's mode is 0.
From a pure statistical point of view, without any knowledge of interest rate, I'd recommend log-normal as in modeling stock prices and inverse-gamma or gamma distribution which are used to model variance or other ...
4
These returns are almost always modeled by finding some fundamental two-sided variable and modeling that. For options, we would model their prices as derivatives -- we would take the log-returns of underlying prices as the fundamental variable, possibly with other models for what would happen to volatilities and the like, and compute the consequences for ...
4
That can be a somewhat difficult question to answer, given that the context may yield different distributions. Nevertheless, I think that you could try to fit the best distribution algorithmically. For instance, lately I found this package at Matlab file exchange:
Finding the best distribution that fits the data
Link
(...) This is where Mike's allfitdist ...
4
It's not possible with a simple linear transformation like the one you mentioned: since scale and thus the distance between mean and median are required to change, either the mean or the median will not be preserved.
Therefore you must use nonlinear transformations, which will complicate quite a bit mantaining skew and kurtosis and imho will not be ...
4
Hi bcf: This is a good question. As you pointed out below,
\begin{align*}
p_0 &= \delta(y-y_0)\\
&=\delta(e^w-y_0).
\end{align*}
Then,
\begin{align*}
p_0 * g &= \int_{-\infty}^{\infty}\delta(e^z-y_0) g(w-z) dz\\
&=\int_{0}^{\infty}\delta(u-y_0) g(w-\ln u) \frac{1}{u}du\\
&= \frac{1}{y_0}g(w-\ln y_0).
\end{align*}
Consequently, your last ...
4
Surely, there is; search for aggregational gaussianity in Google Scholar or ScienceDirect.
In fact, 5 minutes returns are leptokurtic and fat-tailed; then as you increase timeframe, returns become more and more normal. Yearly data is almost normal, if you have enough points.
4
The VaR of level $\alpha$ a loss random variable (the bigger the worse) is the quantity $q$ such that the loss is bigger with probability $1-\alpha$.
Thus we need a $q$ such that
$$
P[L>q] = 1-\alpha,
$$
where we can imagine $\alpha=99\%$ and thus we need the starting point of the $1\%$ tail. Because we have a probability of a loss of size $0$ of $75\%...
4
In this related question How to derive the implied probability distribution from B-S volatilities?, it is shown how to infer the implied probability density of the future prices of a risky asset from a continuum of call prices written on that asset (Breeden-Litzenberger identity).
The developments, which I invite you to read, basically rely on the fact ...
4
Answer
If you assume your returns are independent (yes your models might loosen this assumption) then the two models, $Q_1$ and $Q_2$ assign probability distributions to the returns on any given day, $i$: $q_1^i(r^i)$ and $q_2^i(r^i)$.
Presumably you are interested in the model that can more accurately predict the state of the market over subsequent ...
3
Maybe this could also be a comment but I think an it is not possible to answer this question with a 'yes and here is how you do it'.
It has been tried, e.g. by me for a university research project. In this research we focused primarily on aggregation of returns and the main problem was the tractability of the resulting distributions and expressions, also ...
3
It could be much more simple: if you use the method of moments (MM) then you estimate the mean and the variance and for example the kurtosis of your sample. Then you fit the parameters to these statistics.
Alternatively you use maximum-likelihood (MLE).
For MM: from wikipedia you get the mean and the variance. In your notation you can fit $b = \bar{r}$ so $...
3
the risk neutral drift is needed for pricing of derivatives. For a $100\%$ equity portfolio you can take the real world drift - sometimes a good guess is a drift of zero.
For fixed-income you could do the same and might need more sophistication for the variance term. If you have short-dated bonds then you will need a special model for the pull-to-par.
For ...
3
The pdfs of Student-t distributions have asymptotically Paretian tails, and the tail shape parameter (aka the maximal moment exponent) is equal to the distribution's degrees of freedom parameter. Assuming you have enough observations, you could estimate the Pareto parameter using the so-called Hill method (named after Bruce Hill, 1975). A word of caution: ...
3
What about this sketch of an answer: Let's put $T=1$ in your formula to simplify the notation. Then $Y_b(t)$ is a Brownian bridge where $Y_b(0)=0$ and $Y_b(1)=b$.
This can be written as $Y_b(t) = b\ t + Y_0(t)$, that is to say the standard Brownian bridge (from zero to zero) with an added drift $b\ t$.
The standard Brownian bridge can be written in terms ...
3
My main reference will be "Dan Xu, Christian Beck - Transition from lognormal to chi-square superstatistics for financial time series"
Non-equilibrium statistical mechanics (more specifically, superstatistics) gives some ideas of explaining the relation between time frame and its distribution: "...to regard the time series as a superposition of local ...
3
Your question is not clear. What you might want to say is what distribution should the futures price follow, under the risk-neutral or physical probability measure. In this sense, it will depend on your intention. For potential future exposure, you may want to use the physical measure for the price evolution, while the distribution will depend on your model ...
3
You know that :
$X \sim N(\mu,\sigma^2)$.
$Z = \large\frac{X-\mu}{\sigma}$.
$\text{Var}(Z) = \large\frac{1}{\sigma^2}\text{Var}(X) = \large\frac{1}{\sigma^2}\sigma^2 = 1$.
So that $Z \sim N(0,1)$.
However note that the pdf evaluated for X and Z have different
domains.
The following figure illustrate it :
$X$ is plotted in a) and $Z$ in b)
Their ...
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