# Reason for 0 in discounted stock price process

Let's assume $$dD_t = rD_tdt$$ ($$D_t$$ is Bond Price) and $$dS_t = rS_tdt + σS_tdW_t$$

The reference said $$dD_tdS_t = 0$$

But I don't understand the reason why it is zero.

It said, the Bond Price is deterministic so quadratic stock variation goes to zero. However why the deterministic term makes it zero when it is producted stochastic process?

By the definition of the quadratic covariation

$$\int_0^t dD_u dS_u = [D,S]_t = \lim_{\Vert P\Vert \to 0}\sum_{k=1}^{n}\left(D_{t_k}-D_{t_{k-1}}\right)\left(S_{t_k}-S_{t_{k-1}}\right).$$

We note that:

$$|\sum_{k=1}^{n}\left(D_{t_k}-D_{t_{k-1}}\right)\left(S_{t_k}-S_{t_{k-1}}\right)|\leq \max_{1\leq k\leq n} |S_{t_k}-S_{t_{k-1}}| \left( \sum_{k=1}^{n}|D_{t_k}-D_{t_{k-1}}| \right)$$

Further we note that

$$\max_{1\leq k\leq n} |S_{t_k}-S_{t_{k-1}}| \leq \max_{|u-v|\leq \Vert P\Vert} |S_u -S_v|$$

which will tend to $$0$$ when $$\Vert P\Vert$$ approaches $$0$$, as $$S$$ is a continuous process:

$$\lim_{\Vert P\Vert \rightarrow 0}\max_{|u-v|\leq \Vert P\Vert} |S_u -S_v| = 0 \: \: \: (1)$$

Also, $$\sum_{k=1}^{n}|D_{t_k}-D_{t_{k-1}}| \leq V_t(D),$$

where $$V_t(D)$$, the variation of the process $$D$$ over interval $$[0,t]$$, is finite, as $$D$$ is continuous and deterministic.

Hence the limit above that defines the quadratic variation is $$0$$ (for any $$t$$).

• Thanks for great, detail reply. The one thing makes me confused : In the link, it said 'By the continuity of X, this vanishes in the limit as ${\displaystyle \Vert P\Vert }\Vert P\Vert$ goes to zero.' But I'm not the one who have much experience in math. So Is there any reference or keyword about your reply (in your reply : S 'As partition P gets denser, $max|S_{t_k}-S{t_k-1}|$approaches 0, as S is a continuous process.') that I could search? Apr 6, 2021 at 4:19
• @user13232877 I just edited the text. Statement (1) is the concise way to restate my original clumsy 'dense partitions' statement version.
– ir7
Apr 6, 2021 at 14:52