5

Your equations are flawed. Also there is no expectation if the process $\{r_s\}$ is deterministic. The correct reasoning is, assuming $\{r_s\}$ is stochastic: $$ f(t,u)=-\frac{d}{du}\ln P(t,u)=-\frac{\frac{d}{du}P(t,u)}{P(t,u)}\\ =-\frac{\frac{d}{du}E^Q_t[e^{-\int_t^u r_s ds}]}{P(t,u)} =\frac{E^Q_t[e^{-\int_t^u r_s ds} r_u]}{P(t,u)} =E^Q_t\left[\frac{e^{-\...


3

This is indeed a standard result. You can convince yourself by noticing The bank account grows from 1 at $t=\tau$ to $E\left[\exp(\int_\tau^T r(u)du)|\mathscr{F}_\tau\right]$ at time $T$ The price of a security paying $X$ at time $T$ discounted to $t=\tau$ is then $E\left[X \exp(-\int_\tau^T r(u)du)\right|\mathscr{F}_\tau]$ Hence the price of a credit risk-...


3

Yes, LIBOR rates can be simulated using short rate models. Or rather, Libor rates can be obtained from simulated short rate values. Usually, you have formulas giving you the zero-coupon bond price as a function of the short rate. For affine models for example, this would be of the form: $$P(t, T) = e^{A(t, T) - r(t)B(t,T)}$$ (for example, for the one-factor ...


2

When taking the partial derivative $\frac{\partial}{\partial t}$ in a conditional expectation, not only the parameter $t$ within the expectation needs to be considered, the information set $\mathscr{F}_t$ should also be considered. For this particular question, based on an answer to this question, \begin{align*} P(t, T) = e^{-(T-t)r_t - \int_t^T (T-u)\...


2

In practice, most derivatives traded on Fed Funds rates are linear(i.e. Forwards) rather than non-linear (options and exotics). As such, there has not been a strong case for precise modeling of the full distribution of a Fed Funds rate for a particular day. In contrast , there is a large market for derivatives on 3month USD Libor , which is less sensitive ...


2

I solved from (2) to (4) by myself ! (2) My answer Use the result of (1) with keeping in mind that the following R.H.S is $\mathcal{F}_t $ measurable. \begin{eqnarray} V_t &=& E^{\mathbb{P}} \left[ \exp \left(- \int^T_t r_s ds \right) \cdot ( P(T,S) - K )^+ \middle| \mathcal{F}_t \right] \\ &=& P(t, T) E^{ \tilde{\mathbb{P}} } \left[ ( P(...


2

Let $r(s)$ be the process of a short rate. Then, by risk neutral pricing, $$ P(t,T) = \mathbb{E}^\mathbb{Q}\left[ \exp\left( -\int_t^T r(s)\mathrm{d}s\right) \Bigg| \mathcal{F}_t\right].$$ Thus, the zero-coupon bond is determined completely by the short rate process. Here, $P(t,T)$ denotes the time $t$ price of a zero-coupon bond maturing at time $T$. You ...


1

Treasury / OIS spread is simply the difference between a given Treasury bond's yield (typically the on-the-run Treasuries, like 2y, 5y, etc.) and the fixed rate on an OIS of a similar tenor. If you consider OIS to be a decent proxy for repo rates, the Treasury / OIS spread is a way of gauging how cheap / rich Treasuries are versus their funding. Typically, ...


1

(Cumulative Integration Formula Replacing $du$ and $dB_s$) I have developed formulas to solve this by myself! \begin{eqnarray} \int^t_0 \int^u_0 dB_s \ du &=& \int^t_0 \int^u_s du \ dB_s \\ \int^T_t \int^u_0 dB_s \ du &=& \int^T_0 \int^u_s du \ dB_s - \int^t_0 \int^u_s du \ dB_s \end{eqnarray} Therefore, we can use the following ...


1

(My answer) the Vasicek Bond Price and its Forward Price Recall the result of Exercise 5.2.(1) or Exercise 4.5.(10). \begin{eqnarray} P(t, T) &=& E \left[ \exp \left( - \int^T_t r_u du \right) \middle| \mathcal{F}_t \right] \\ &=& E \left[ \exp \left( - \int^T_t \left( e^{-bu} r_0 + \sigma \int^u_0 e^{-b(u-s)} dB_s \right) du\right) ...


1

I solved by myself. The following is this solution. Let $T-t=s$, one reaches the following equation. \begin{eqnarray} B'(s) + \beta B(s) + \frac{1}{2} \sigma^2 B(s)^2 =1 \end{eqnarray} One finds out it is the Riccatti equation because of $A(s)=0$. Therefore, one reaches the following equation. \begin{eqnarray} B' = - \frac{1}{2} \sigma^2 B^2 - \...


1

I found the answer to my question! It consists of separating the terms $1$ and $e^{-a(T-u)}$ from $B(u,T)$ and isolating the $e^{aT}$ term from the integral and it's straight forward then! All in all it's the form of $B(u,T)$ that makes it possible to state such a formula. PS: In order to continue the demonstration, we need to derive $-\frac{\partial^2}{\...


1

The problem should go away if you simulate $r_t$. Ho Lee should work for the function of the form you assumed: $P(0,T)=e^{-aT^2-bT}=e^{-(aT+b)T}$ The problem with your simulation is that the forward rate, as you correctly derived, is as follows: $f(0,T)=2aT+b$ So when you take the derivative to calculate $\theta$, you lose b. But remember the short rate ...


1

The main thing we want is the $P(t,T)$ function. In the short rate model, we model the system as an instantaneous short rate variable which evolves stochastically. Different models assign different dynamics to the short rate (mean reversion, constant or stochastic vol, etc), but they all assume that $P(t,T)$ is the expectation of the integral of the ...


1

Although it's been a long time this question has been asked, I'd like to propose an answer in case someone was looking for the same thing. First, I think there's a confusion between $P(t,T)$ and $DF(t,T)$. The former is the $t-$price of a contract paying $1$ unit of currency at date $T$ while the later is the (stochastic) discount factor at $t$ for flows ...


1

CIR can be used to simulate paths, although forecasting with a model of that class is a bit unintuitive. Why? The results are highly dependent on the stochastic parameters of the equation. So, let's say you obtained a calibrated model and simulated it 5 times. You would (most likely) get very different results with the same model but with different random ...


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