Conditional Moments of the CIR Variance Process
Summary
The document derives the conditional expectation and variance of a square-root mean-reverting process, then applies the results to the variance process used in a Heston model. It first takes the conditional expectation of the stochastic differential equation; the Brownian integral has zero conditional mean, leaving a linear integral equation. Solving it gives a conditional mean that decays from the current state toward the long-run level.
For the second moment, the derivation applies Ito’s formula to the squared process, takes conditional expectations, and solves the resulting differential equation using the first-moment result. The resulting expression gives the conditional variance over a future interval. The method is useful for understanding variance dynamics and for calculations involving discretization or simulation. The source contains a notation slip in its final variance line, labeling the variance of the squared process even though the expression is obtained by subtracting the squared conditional mean of the process itself; parameters must also be mapped carefully to the Heston notation.
Key ideas
- The Heston variance process is a special case of the Cox-Ingersoll-Ross square-root process.
- Its conditional mean reverts exponentially from the current variance toward the long-run level.
- Applying Ito’s formula to the squared state yields an equation for its conditional second moment.
- Subtracting the squared conditional mean from the second moment gives the conditional variance.
- The source’s final variance label appears inconsistent with the quantity calculated.
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# Distribution in Heston
# Distribution in Heston
$$dV_t=-k(V_t-1)dt+ \epsilon\sqrt{V_t}dW_t$$ $W_t$ is wiener process and the rest is just some parameters.
For $T_{i+1}>T_{i}$ how do I find the expectation and variance of $V_{T_{i+1}}$ conitional to $V_{T_i}$?
## Answer by hypernova (score 3, accepted)
https://quant.stackexchange.com/a/39582
As per @Quantuple, we may consider a more general Cox-Ingersoll-Ross model $$ {\rm d}r_t=a\left(b-r_t\right){\rm d}t+\sigma\sqrt{r_t}{\rm d}w_t. $$ Integrating this equation on $t\in\left(u,v\right)\subseteq\mathbb{R}^+$ yields \begin{align} r_v-r_u=\int_u^v{\rm d}r_t&=\int_u^v\left[a\left(b-r_t\right){\rm d}t+\sigma\sqrt{r_t}{\rm d}w_t\right]\\ &=ab\left(v-u\right)-a\int_u^vr_t{\rm d}t+\sigma\int_u^v\sqrt{r_t}{\rm d}w_t. \end{align} Acting $\mathbb{E}\left(\cdot|r_u\right)$ on both sides of this equality gives \begin{align} \mathbb{E}\left(r_v|r_u\right)-r_u&=\mathbb{E}\left(r_v-r_u|r_u\right)=\mathbb{E}\left[ab\left(v-u\right)-a\int_u^vr_t{\rm d}t+\sigma\int_u^v\sqrt{r_t}{\rm d}w_t\Bigg|r_u\right]\\ &=ab\left(v-u\right)-a\mathbb{E}\left(\int_u^vr_t{\rm d}t\Bigg|r_u\right)+\sigma\mathbb{E}\left(\int_u^v\sqrt{r_t}{\rm d}w_t\Bigg|r_u\right)\\ &=ab\left(v-u\right)-a\mathbb{E}\left(\int_u^vr_t{\rm d}t\Bigg|r_u\right)\\ &=ab\left(v-u\right)-a\int_u^v\mathbb{E}\left(r_t|r_u\right){\rm d}t. \end{align} Denote $f(t)=\mathbb{E}\left(r_t|r_u\right)$ for $t\ge u$, and the last equation is equivalent to, for all $v\ge u$, $$ f(v)=r_u+ab\left(v-u\right)-a\int_u^vf(t){\rm d}t. $$ Regard this as an integral equation with respect to $v$, and its respective differential equation with initial condition reads \begin{align} f'(v)&=ab-af(v),\\ f(u)&=r_u. \end{align} This system immediately leads to
> $$ \mathbb{E}\left(r_v|r_u\right)=f(v)=r_ue^{-a\left(v-u\right)}+b\left(1-e^{-a\left(v-u\right)}\right). $$
Based on the result above, we may figure out $\mathbb{E}\left(r_v^2|r_u\right)$ as follows. Thanks to Ito's formula, the Cox-Ingersoll-Ross model gives \begin{align} {\rm d}\left(r_t^2\right)=2r_t{\rm d}r_t+{\rm d}\left<r\right>_t&=2ar_t\left(b-r_t\right){\rm d}t+\sigma r_t\sqrt{r_t}{\rm d}w_t+\sigma^2r_t{\rm d}t\\ &=\left[\left(2ab+\sigma^2\right)r_t-2ar_t^2\right]{\rm d}t+\sigma r_t\sqrt{r_t}{\rm d}w_t. \end{align} Again, Integrating this equation on $t\in\left(u,v\right)\subseteq\mathbb{R}^+$ yields $$ r_v^2-r_u^2=\left(2ab+\sigma^2\right)\int_u^vr_t{\rm d}t-2a\int_u^vr_t^2{\rm d}t+\sigma\int_u^vr_t\sqrt{r_t}{\rm d}w_t. $$ Acting $\mathbb{E}\left(\cdot|r_u\right)$ on this equality yields $$ g(v)-r_u^2=\left(2ab+\sigma^2\right)\int_u^vf(t){\rm d}t-2a\int_u^vg(t){\rm d}t, $$ where $f$ agrees with the notation from above, while $g(v)=\mathbb{E}\left(r_t^2|r_u\right)$. Note that its respective differential equation reads \begin{align} g'(v)&=\left(2ab+\sigma^2\right)f(v)-2ag(v),\\ g(u)&=r_u^2. \end{align} Thanks to this system, we could eventually figure out
> \begin{align} \text{Var}\left(r_v^2|r_u\right)&=\mathbb{E}\left(r_v^2|r_u\right)-\mathbb{E}^2\left(r_v|r_u\right)=g(v)-f^2(v)\\ &=\frac{\sigma^2}{a}\left[r_ue^{-a\left(v-u\right)}\left(1-e^{-a\left(v-u\right)}\right)+\frac{b}{2}\left(1-e^{-a\left(v-u\right)}\right)^2\right]. \end{align}
The answer to your original question could be obtained by taking $a=k$, $b=1$, $\sigma=\epsilon$, $u=T_i$ and $v=T_{I+1}$.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.