# What is time 0 price of Libor starting t for the period $t$ to $t+\delta t$

I was asked this in an interview.

Let $L_0[0,t]$ denote the time 0 price of Libor for period $0$ to $t$.

Let $L_0[t,t+\delta_t]$ denote the time 0 price of Libor for period $t$ to $t+\delta t$

$(1+L_0[0,t])^{-1}$ dollar today is worth $1$ dollar at time $t$ hence $1+L_0[t,t+\delta_t]$ dollar at time $t+\delta t$

So the correct answer should be $(1+L_0[0,t])^{-1}-(1+L_0[0,t+\delta t])^{-1}$ by discounting the extra 1 dollar back to time 0.

I understand this, but I thought

$1$ dollar today is worth $1+L_0[0,t]$ dollar at time $t$, hence $(1+L_0[0,t])(1+L_0[t,t+\delta_t])$ dollar at time $t+\delta t$

That should be the same as investing 1 dollar for period 0 to $t+\delta t$. This should be $1+L_0[0,t+\delta t]$. Equating these give a different answer.

$(1+L_0[0,t])(1+L_0[t,t+\delta_t])= 1+L_0[0,t+\delta t]$

This gives $L_0[t,t+\delta_t] = (1+L_0[0,t+\delta t])(1+L_0[0,t])^{-1} - 1$

This is out by a factor of $1+L_0[0,t+\delta t]$. why is that the case? what did I miss?

The answer $$(1+L_0[0,t])^{-1}-(1+L_0[0,t+\delta t])^{-1}$$ is incorrect.

Note that, since $(1+L_0[0,t])^{-1}$ dollar toady worth $1+L_0[t, t+\delta t]$ at time $t+\delta t$, and $(1+L_0[0,t+\delta t])^{-1}$ dollar today worth 1 dollar at time $t+\delta t$, therefore, $$(1+L_0[0,t])^{-1}-(1+L_0[0,t+\delta t])^{-1}$$ dollar today worth $L_0[t, t+\delta t]$ at time $t+\delta t$, then \begin{align*} (1+L_0[0,t])^{-1}-(1+L_0[0,t+\delta t])^{-1} &= L_0[t, t+\delta t] P_0(t+\delta t)\\ &=L_0[t, t+\delta t] (1+L_0[0,t+\delta t])^{-1}. \end{align*} That is, \begin{align*} L_0[t, t+\delta t] = (1+L_0[0,t+\delta t])(1+L_0[0,t])^{-1} - 1, \end{align*} which is the same as your answer in your second part.

EDIT:

Here, it is assumed that $L(0, t) = \frac{1}{P_0(t)}-1$, where $P_0(t)$ is the price at time $0$ of a zero-coupon bond with maturity $t$ and unit face value.

• sorry, what is your $P_0(t+\delta t)$? Oct 12, 2016 at 14:58
• The price at tome 0 for a zero-coupon bond with maturity $t+\delta t$. Here, it is assumed that $L_0(0, t+\delta t) = \frac{1}{P_0(t+\delta t)}-1$. Oct 12, 2016 at 15:00
• The two answers are out by a factor $(1+L_0[0,t+\delta t])^{-1}$. So i think he asked for the time zero value of a payment of $L_t[t,t+\delta t]$, which is what he said the answer was for. I now finally understand what is going on. Looks like I won't be getting the job. :) Oct 14, 2016 at 13:22
• @Lost1: The expected Libor payment at $t+\delta t$ is indeed $(1+L_0[0,t])^{-1}-(1+L_0[0,t+\delta t])^{-1}$, but the time $0$ value $L_0(t, t+\delta t)$ is not, for which you second derivation is correct. Oct 14, 2016 at 13:32
• @Lost1: additionally, the time 0 value $L_0(t, t+\delta t)$ is the expected Libor payment at $t+\delta t$ discounted back to time $0$. Oct 14, 2016 at 13:47