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Vasicek Short Rate Integrals over a Future Time Interval

Article Quant Q&A · Author: user1171853
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Summary

The document asks how to calculate the conditional expected value and variance of the accumulated short rate between two future times under the Vasicek model. It begins with formulas for the integral from time zero to a given horizon, then obtains the interval expectation by subtracting the two cumulative expectations. This produces an expression involving the model’s initial short rate, mean reversion level and speed, and the interval endpoints.

For variance, the author proposes subtracting the cumulative variances. That step is not generally valid: the two cumulative integrals overlap, so the variance of their difference also depends on their covariance. The document does not supply a verified interval variance result or an answer to that issue. Its value is as a focused example of why expectations can be differenced linearly while variances require dependence terms.

Key ideas

  • The question concerns moments of an accumulated Vasicek short rate over an interval after time zero.
  • The interval expectation is formed by subtracting cumulative expectations at the two endpoints.
  • The proposed variance subtraction omits the covariance between cumulative integrals.
  • The text poses the variance calculation but does not provide a corrected result.

Tags

Fixed incomeStatisticsDerivatives pricing
Full text
# Expected value and variance of the short rate under the Vasicek model


# Expected value and variance of the short rate under the Vasicek model












Would be grateful for any assistance. Below are the expected value and variance of the integral of the short rate under the Vasicek model (https://www.researchgate.net/publication/41448002):

$E\left[ \int_{0}^{t}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\frac{(r_{0}-b)}{a}(1-e^{-at})+bt$ $Var\left[ \int_{0}^{t}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\frac{\sigma^{2}}{2a^{3}}(2at-3+4e^{-at}-e^{-2at})$

But what if I would like to find the following:

$E\left[ \int_{t_{1}}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{} \\ Var\left[ \int_{t_{1}}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{} \\ where \;\;0\lt t_{1}\lt t_{2}$

My question is, can I simply rewrite the above expressions as:

$E\left[ \int_{t_{1}}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=E\left[ \int_{0}^{t_{2}}r(u)du - \int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{} = \frac{(r_{0}-b)}{a}(e^{-at_{1}}-e^{-at_{2}})+b(t_{2}-t_{1})$

$Var\left[ \int_{t_{1}}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{} = Var\left[ \int_{0}^{t_{2}}r(u)du-\int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\\=Var\left[ \int_{0}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}+Var\left[ \int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}-2Var\left[ \int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\\=Var\left[ \int_{0}^{t_{2}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}-Var\left[ \int_{0}^{t_{1}}r(u)du|\mathcal{F_{0}}\right]\mathcal{}=\\= \frac{\sigma^{2}}{2a^{3}}(2a(t_{2}-t_{1})+4(e^{-at_{2}}-e^{-at_{1}})-e^{-2at_{2}}+e^{-2at_{1}})\\\\$

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.