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I have an unobservable stochastic quantity $\lambda(t)$, which I analytically know the variance of, that is

$$\text{Var}(\lambda(t))= \frac{\theta \sigma^2}{2\kappa}$$

My goal is to get an estimate of $\sigma^2$.

I can observe S and K historically at times $t=1,2,..$, and know that the following approximately holds

$$ \lambda(t) \approx S(t)+K(t)$$

Does it make sense to simply take the sample variance of the time series $(S(t)+K(t))$, let's call it $\hat{\sigma}^2_S$, and say that a reasonable estimate of $\sigma^2$ is $2 \kappa\frac{\hat{\sigma}^2_S}{\theta}$?

It should be mentioned that the distribution of $\lambda$ is not pretty, but I have the mean and variance.

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    $\begingroup$ yes it is correct if your $\approx$ is a "reasonable" estimate. $\endgroup$ – MJ73550 Aug 5 '16 at 11:56
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    $\begingroup$ it is correct unless S(t) and K(t) are correlated because in this case var(S+K) is not equal to var(S) + var(K). $\endgroup$ – Malick Aug 5 '16 at 22:19
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    $\begingroup$ Does it really matter, if S and K are correlated? If you have a sample of S(t) + K(t) you can estimate the variance of that sample regardless of the distribution of S and K, right? $\endgroup$ – Ami44 Aug 7 '16 at 14:58
  • $\begingroup$ What is the process in question? Where did you get the formula for the variance? This is interesting for the background of the question. $\endgroup$ – SRKX Jan 5 '17 at 6:44
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If $$ \Lambda = S + K $$ then you can look at samples of $S+K$ and estimate the variance of $\Lambda$ by the variance of $S+K$. If $\theta$ and $\kappa$ are known constants then you can do the algebra to derive $\sigma^2$.

The question is how the above equality holds. If it holds almost surely, then you are done. If it holds in probability then you are done too.

Maybe it helps to look at

$$ \Lambda_t = S_t + K_t + \epsilon_t $$ and model that resiudal $\epsilon$.

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$$ \text{Var}(dL) = \text{Var}(dK) + \text{Var}(dS) + \text{cov}(dS, dK) $$ I hope u have some simultaneous observations on $dK$ and $dS$ to estimate the covariance or a good assumption about it.

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    $\begingroup$ If he has a simultaneous sample of $S$ and $K$, them he can compute a sample of $\lambda$ and does not need to care about the covariance... $\endgroup$ – SRKX Jan 5 '17 at 6:46

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