Why Discount-Factor Distributions Depend on Integrated Short Rates
Summary
The document asks how to derive the distribution of a discounted LIBOR-related quantity when the short rate follows a CIR model. It starts from a discounted payoff involving the exponential of the negative integral of the short rate, then tries to convert its cumulative probability into a statement about the terminal short rate. The key issue is that a CIR marginal distribution for the rate at one time does not by itself give the distribution of the time integral of the rate.
The proposed differentiation step is invalid: differentiating an inequality that depends on the threshold does not preserve the probability event or turn the integral into its endpoint value. The LIBOR term may also be stochastic and dependent on the discount factor, so its joint behavior matters. A proper calculation needs the distribution of the integrated CIR rate and, where relevant, its dependence with the LIBOR rate; affine term structure methods can characterize related Laplace transforms. The document provides no derivation of that distribution or numerical evidence, and its attempted endpoint reduction does not establish a CDF.
Key ideas
- A discount factor driven by a short rate depends on the integral of the rate over time.
- The marginal distribution of a CIR short rate does not determine the distribution of its time integral.
- Differentiating an inequality does not convert an integral event into an event involving only the terminal rate.
- If the LIBOR rate is random, its dependence with the discount factor must also be considered.
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Full text
# How can I find the distribution function of the following random variables?
# How can I find the distribution function of the following random variables?
Suppose that the random variables $Z_i$ are defined as follows:
\begin{equation} Z_i = D(0, t_i)(R_{i-1} +c)\Delta N, \end{equation} where $D(0, t_i)= \exp\{-\int_{0}^{t_i} r_u du\}$ for which $r_u$ follows a CIR model, $R_{i-1}$ stands for the forward LIBOR rate, and $c$, $\Delta$, $N$ are positive constants. I am looking for the distribution function of $Z_i$. However, we already know the fact that $r_u$ has a non-central Chisquare distribution. My main issue is how to get rid of the integral on the left side of the inequality below. I think there is something wrong. I am not sure whether we need to take into account a specific rule when we want to take differential of both sides of an inequality. Let me share my solution. For $y> 0$, we have that
\begin{equation} \mathbb{P}(Z_i \leq y) = \mathbb{P}\Big(D(0, t_i)(R_{i-1} +c)\Delta N \leq y\Big) = \mathbb{P}\Big(D(0, t_i) \leq \frac{y}{(R_{i-1}+c)\Delta N}\Big) = \mathbb{P}\Big(\exp\{-\int_{0}^{t_i} r_u du\} \leq \frac{y}{(R_{i-1}+c)\Delta N}\Big) = \mathbb{P}\Big(\int_{0}^{t_i} r_u du \geq \log\Big(\frac{(R_{i-1}+c)\Delta N}{y}\Big)\Big) = \mathbb{P}\Big(d\left[\int_{0}^{t_i} r_u du\right]\geq d\left[\log\Big(\frac{(R_{i-1}+c)\Delta N}{y}\Big)\right]\Big) = \mathbb{P}\Big(r_{t_i}dt_i\geq d\left[\log\Big(\frac{(R_{i-1}+c)\Delta N}{y}\Big)\right]\Big) = \mathbb{P}\Big(r_{t_i}\geq \frac{d}{dt_i}\left[\log\Big(\frac{(R_{i-1}+c)\Delta N}{y}\Big)\right]\Big) = \mathbb{P}\Big(r_{t_i}\geq \frac{R^{\prime}_{i-1}}{R_{i-1}+c}\Big) \end{equation} where the last line is obtained by knowing ${[\log(f)]}^{\prime} = \frac{f^{\prime}}{f}$. Note here that the LIBOR rate is defined as follows: \begin{equation} R_{i-1} = \frac{1}{t_i - t_{i-1}}\Big(\frac{P(t_{i-1}, t_{i-1})}{P(t_{i-1}, t_i)} -1\Big) = \frac{1}{\Delta}\Big(\frac{1}{P(t_{i-1}, t_{i})} -1\Big) \end{equation} where $P(t_{i-1}, t_i)$ denotes the zero-coupon bond price issuing at time $t_{i-1}$ and maturing at time $t_i$ with following solution
\begin{equation} P(t_{i-1}, t_i) = A(t_{i-1}, t_i)\exp\Big\{-B(t_{i-1}, t_i)r(t_{i-1})\Big\} \end{equation}
, $t_{i} - t_{i-1} = \Delta$, and finally the notation $d[.]$ represents the derivative of a given term. Moreover, $A(., .)$ and $B(., .)$ are deterministic functions of times $t_{i-1}$ and $t_i$.
As you can see, there is no $y$ left at the end. I doubt that the final result is showing a distribution function because we have a constant, not a function of $y$.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.