any guidance on how to prove the linearity of quadratic variation would be greatly appreciated.

Denote quadratic variation of a continuous stochastic process $$X_t$$ as $$\left[X,X\right]_t = p-\lim_{max_{i}|t_{i+1}-t_i| \to 0}\sum_{i=0}^{n-1}{|X_{t_{i+1}}-X_{t_i}|}^2$$ and given that the quadratic covariation of two continuous square integrable martingales $$X_t$$ and $$Y_t$$ is defined as \begin{align} \left[X,Y\right]_t &= \frac{1}{4}\bigl(\left[X+Y,X+Y\right]_t - \left[X-Y,X-Y\right]_t\bigr) \\ &=\frac{1}{2}\bigl(\left[X+Y,X+Y\right]_t-\left[X,X\right]_t - \left[Y,Y\right]_t\bigr) \end{align}

show that $$\left[aX+bY,Z\right]_t = a\left[X,Z\right]_t + b\left[Y,Z\right]_t$$

EDIT: I am mainly hesitant because of the modulus and square exponent. Perhaps, let's try to show a simplier example: $$[X+Z,Z] = [X,Z]+[Z,Z]$$ (I drop off the index)

LHS: $$[X+Z,Z]=\frac{1}{2}[X+2Z,X+2Z]-\frac{1}{2}[X+Z,X+Z]-\frac{1}{2}[Z,Z]$$

RHS: \begin{align} [X,Z]+[Z,Z] &= \frac{1}{2}([X+Z,X+Z]-[X,X]-[Z,Z]) + [Z,Z] \\ &= \frac{1}{2}[X+Z,X+Z] - \frac{1}{2}[X,X] + \frac{1}{2}[Z,Z] \end{align}

Should I be approaching this problem as such?

Thanks!

• Denote $Q$ the quadratic variation. I think you should first prove that $Q(X+Y)+Q(X-Y) = 2Q(X)+2Q(Y)$. – Jakobian Jul 25 at 19:08
• @Jakobian I don't quite see how the above property helps? – Galvin Ng Jul 27 at 4:44
• What have you tried? It's really not that difficult, just a matter of using the very definitions of the objects appearing in the questin. – saz Jul 27 at 13:19
• Well, I've thought about this in the following way. Quadratic variation looks a bit like a squared norm, so why not prove the rectangle equality to prove it's given by a scalar product? – Jakobian Jul 27 at 16:23
• @Jakobian would you like to take a look at the answer I put up? Would you agree with my approach? I realized my initial doubt of using the triangle inequality was a mere scare. Just a tedious computation... Thanks! – Galvin Ng Jul 28 at 18:14

One approach would be to first show $$[f,g]_t = \lim_{\delta_n \to \,0}\sum_{i=0}^{n-1}(f_{t_{i+1}} - f_{t_{i}})(g_{t_{i+1}} - g_{t_{i}})$$ then we get$$[\alpha f+ \beta g,h]_t = \lim_{\delta_n \to \,0}\sum_{i=0}^{n-1}\left[(\alpha f + \beta g)_{t_{i+1}} - (\alpha f + \beta g)_{t_{i}})(h_{t_{i+1}} - h_{t_{i}})\right]$$ and $$\alpha[f,h]_t = \alpha\left[\lim_{\delta_n \to \,0}\sum_{i=0}^{n-1}(f_{t_{i+1}} - f_{t_{i}})(h_{t_{i+1}} - h_{t_{i}})\right]$$ $$\beta[g,h]_t = \beta\left[\lim_{\delta_n \to \,0}\sum_{i=0}^{n-1}(g_{t_{i+1}} - g_{t_{i}})(h_{t_{i+1}} - h_{t_{i}})\right]$$
• It's an immediate consequence of your definition of the covariation $$[X,Y]_t = \frac{1}{4} ([X+Y,X+Y]_t- [X-Y,X-Y]_t).$$ Use the definition of the quadratic variation (for the right-hand side) and expand the squares. – saz Jul 29 at 18:50
• Yes. I did expand the terms for $[\alpha X + \beta Y, Z]$ and obtained $\lim \sum \bigl[(\alpha X_{t_{i+1}} + \beta Y_{t_{i+1}}) - (\alpha X_{t_{i}} + \beta Y_{t_{i}})\bigr](Z_{t_{i+1}} - Z_{t_i})$ but I realized that is similar to the first line of the answer I posted. As such, I decided to start off (in my answer) as shown above. – Galvin Ng Jul 30 at 5:17