4

The Exponentially Weighted Moving Average (EWMA for short) is characterized my the size of the lookback window $N$ and the decay parameter $\lambda$. The corresponding volatility forecast is then given by: $$ \sigma_t^2 = \sum_{k = 0}^N \lambda^k x_{t-k}^2 $$ Sometimes the above expression is normed such that the sum of the weights is equal to one. ...


3

Look at the infinitesimal version of the change in variance: $$ d\sigma^2 = 2\sigma d\sigma + (d \sigma)^2 $$ The Ito term $(d\sigma)^2$ is non-zero for stochastic processes, and is of order $dt$, but if we ignore that then we get the approximate relation $$ d\sigma^2 \approx 2 \sigma d\sigma $$ which is where the factor $2 \sigma$ comes from in the ...


3

Forget for a moment that your option is delivering the immediate entrance in a swap (if the swaption is physically settled) or the cash amount of the swap (if the swaption is cash-settled), as your question doesn't depend on this fact, and take a "general" 1Y option. Your today's (date $t_0$) cube loses the "swap tenor dimension" and becomes a today's ...


2

Let $X_1$ and $X_2$ be your two assets and $C$ your financial product. For now we only assume products which are a linear combination of $X_1$ and $X_2$ with no shorting allowed, hence: $$\begin{align} & C=\alpha X_1 + (1-\alpha)X_2 \\ & 0\leq \alpha \leq 1 \end{align}$$ Letting $\rho$ be the correlation between returns $r_1$ and $r_2$, we have: $$\...


2

Throw in correlation as the additional variable. Similarly, volofvol could be another candidate to play with. A spread option could have a larger volatility than either of the two underliers.


2

$S_t$ is log-normal, so indeed its variance will be different. $\sigma ^2$ is, however, the variance of the returns $\log S_t$ per unit of time since $$\log S_t \sim N\left(\log S_0 + \left(r-\frac 12 \sigma ^ 2\right) t, \sigma ^2 t\right)$$


1

One of the main features of the Hull-White model is that it matches the market at $t = 0$. This means that at $t = 0$, not only does the zero coupon bond prices (starting from zero) not depend on the volatility, but neither do they depend on the mean reversion level. These prices depend only on the zero curve observed in the market. Of course, this is not ...


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