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

Question

I was asked this in an interview.

The correct answer, I was told, follow from this argument

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.

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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?

Answer 1

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.