I'm diving into financial mathematics and have calculated a matrix that gives me the daily change of four given securities:

The start of it looks like this:

enter image description here

My Tutorial is now calculating the covariance of a stock over a year as $250\ Cov(d_1, d_2)$, where $d_x$ is the daily change of a stock.

250 is the number of trading days on my stock exchange.

So, for example, the iROB and DAX have a daily Covariance of $Cov(iROB, DAX) = -0.000173$ and therefore a yearly one of $250\ Cov(iROB, DAX) = -0.043201$.

This feels to me like comparing apples and oranges. Why can I scale the covariance up by a factor of days in order to get the yearly Covariance?


This is derived under assumption of Wiener process, more simply put if you have random independent identically normally distributed daily increments then covariance (and variance as well) simply adds up as time passes.

Compare this with:

Suppose: $$ Z=X+X' \quad X,X'\sim \mathcal{N}_{0,1} $$ Then: $$ Z \sim \mathcal{N}_{0,2} $$ Which you can easily verify using linearity of expectation as: $$\mathbb{E}[Z^2]=\mathbb{E}[(X+X')(X+X')]=\mathbb{E}[X^2]+\mathbb{E}[{X'}^2]+\mathbb{E}[XX']$$ And $\mathbb{E}[XX']=0$ as increments are independent.

  • $\begingroup$ Thanks, can you elaborate on what's $Z, X, X'$ and $\mathcal{N}_{0,1}$? $\endgroup$ – shredding Feb 20 at 10:06
  • $\begingroup$ $Z,X,X'$ are random variables - they represent 2-day change, 1-day change, 1-day change respectively. $\mathcal{N}_{0,1}$ is standard normal distribution (cf. wiki Normal distribution) $\endgroup$ – Radost Feb 20 at 10:12
  • $\begingroup$ Thanks, I will read into wiener process and report back! $\endgroup$ – shredding Feb 20 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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