1
$\begingroup$

I want to determine the kurtosis of a straddle. My question is closely related with the following topic here. According to the following paper of Ben-Meir and Schiff (2012) the expected value of a call is equal to

enter image description here

where

enter image description here

The variance of the call is

enter image description here

Following the standard definition of kurtosis I can write:

enter image description here

Similar, I can write the same for the puts:

enter image description here

Is it correct to assume that:

I want to calculate the kurtosis of a straddle

$\endgroup$
1
  • $\begingroup$ No it is not. For the same reason you cannot assume the variance of a portfolio equals the sum of the of the individual variances. And it's even weird to talk about variance of a call/putinstead of variance of the return of a call/put. This may help: quant.stackexchange.com/questions/2273/… $\endgroup$
    – phdstudent
    Jun 23, 2020 at 15:37

1 Answer 1

2
$\begingroup$

Even if you assume null cokurtosis terms, your equality is still off:

\begin{align} \operatorname{Kurt}[X+Y] = {1 \over \sigma_{X+Y}^4} \big( & \sigma_X^4\operatorname{Kurt}[X] + \sigma_Y^4\operatorname{Kurt}[Y] \big). \end{align}

Note that you need $\sigma_{X+Y}^2$. You already have $\sigma_X^2$ and $\sigma_Y^2$ (computed in the paper).

Full formula is:

\begin{align} \operatorname{Kurt}[X+Y] = {1 \over \sigma_{X+Y}^4} \big( & \sigma_X^4\operatorname{Kurt}[X] + 4\sigma_X^3\sigma_Y\operatorname{Cokurt}[X,X,X,Y] \\ & {} + 6\sigma_X^2\sigma_Y^2\operatorname{Cokurt}[X,X,Y,Y] \\[6pt] & {} + 4\sigma_X\sigma_Y^3\operatorname{Cokurt}[X,Y,Y,Y] + \sigma_Y^4\operatorname{Kurt}[Y] \big). \end{align}

$\endgroup$
1
  • $\begingroup$ Thanks do you think it is reasonable to assume null cokurtosis terns? $\endgroup$
    – HJA24
    Jul 1, 2020 at 5:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.