Why a Deterministic Bond Price Has Zero Covariation with a Continuous Stock
Summary
The document explains why the quadratic covariation between a deterministic bond price and a continuous stochastic stock price is zero. It considers a bond evolving deterministically and a stock following a diffusion, then uses the partition-sum definition of quadratic covariation to show the result.
Each sum of paired increments is bounded by the largest stock-price move over a partition interval multiplied by the bond’s total variation. As the partition becomes finer, continuity makes the largest stock increment tend to zero, while the deterministic bond path has finite variation over the time interval. Their product therefore tends to zero. This argument supports the zero cross-variation term used in stochastic calculus; it relies on continuity of the stock path and finite variation of the bond path, and the document does not cover discontinuous price processes.
Key ideas
- Quadratic covariation is defined as a limit of sums of paired increments over finer partitions.
- The sum is bounded using the largest stock increment and the bond path’s total variation.
- Continuity makes the largest stock increment vanish as the partition mesh shrinks.
- A deterministic continuous bond path has finite variation, so its covariation with the continuous stock path is zero.
- The argument’s stated assumptions do not cover discontinuous price processes.
Tags
Full text
# Reason for 0 in discounted stock price process
# 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?
reference : https://www.youtube.com/watch?v=TPxnnRYWst8 (Quantpie youtube)
## Answer by ir7 (score 6, accepted)
https://quant.stackexchange.com/a/63229
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 $t$).Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.