# Tag Info

5

I'm not sure I understand, but if you want to compute the variance of $exp(X)$, where $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$, that variance is (from Wikipedia): $$\left(\exp{(\sigma^2)} - 1\right) \exp{(2\mu + \sigma^2)}$$

5

You're forgetting that -2.52 is still in natural logarithm terms. So the correct answer is 2.71828183 raised to the -2.52 power which equals 0.08. Your ending portfolio value is 8% of what it was a year ago.

5

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. ...

4

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

4

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 ...

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

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 $... 3 The distribution of the log of a stock price in n days is a normal distribution with mean of$\log(current_price)$and standard deviation of$volatility*\sqrt(n/365.2425)$if you're using calendar days, and assuming no dividends and 0% risk-free interest rate. Note that the standard deviation is independent of the current_price: if$\log(current_price)$... 3 Options on interest rates futures in the listed markets are always traded 1-yield (100-yield) just like the futures which are traded 1-yield. So negative rates aren't an issue and its always black volatility. In the OTC market, both normal and black volatility are quoted, but the common practice is to use black volatility is what is way more frequently used.... 2 if they are stocks, this problem is called pricing a Margrabe option and it is generally solved by change of numeraire. Take$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 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$S_t$is stochastic process and follow geometric Brownian motion with following SDE: $$dS_t=\mu S_t dt + \sigma S_t dW_t$$ then$S_Tfollows 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-... 1 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)... 1 In that case, the problem becomes a non-trivial stopping time problem. Consider a filtered probability space (\Omega, \mathcal{F}, \mathbb{P}) equipped with the natural filtration of a standard Brownian motion W_t^\mathbb{P}. Assuming a geometric Brownian motion for the underlying asset, one gets S_t = S_0 \exp\left((\mu-\frac{1}{2}\sigma^2)t + \... 1 The formula in your skew function is one of skew normal distribution. That distribution has a limit on skew parameter, while in the real world there is no such limit. From personal experience, few years ago I tried doing exactly what you described in your question. After comparing skew normal distribution on SPX with the real world, I concluded that there ... 1 This is a good question which I got stuck on as well. SupposeX$is lognormal defined as$X\sim \log \mathcal{N}(\mu, \sigma^2)$. With this notation we mean that if we write$X = e^Z$, then$Z$follows a normal distribution with mean$\mu$and variance$\sigma^2$. The mean and variance of$X$are then$\mu_X = e^{\mu + \frac{1}{2}\sigma^2}$and$\sigma_X^...

1

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.

1

Well, log-normality does not allow you to have negative earnings and companies do have negative earnings. I suggest you to download the earnings data and perform a Jarque-Bera test for normality.

1

You know that Brownian motion {W(t)} is a stochastic process with the following properties: (Independence of increments) W(t) − W(s) , for t > s , is independent of the past, that is, of W(u) , 0 ≤ u ≤ s, or of $F_s$ , the σ-field generated by W(u), u ≤ s. (Normal increments) W(t) − W(s) has Normal distribution with mean 0 and variance t − s. This implies ...

1

Your formula, as it stands, is incorrect, at least is if $E$ means the "expected value under real-world probabilities". I wrote a blog post explaining the basic rationale behind risk-neutral pricing where you will see that if the Fundamental Theorem of Asset Pricing theorem holds, you can write: Let $X_t=S_{1,t}-S_{2,t}$ $$e^{-rt} X_t = \mathbb{E}_\mathbb{... 1 One way to start thinking about this is to work out a couple of Discrete versions of Ito's lemma Øksendal (6th edition) Example 3.1.9: almost surely,$$ 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 ...

1

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.

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