# Floating leg of a standard swap still has a value at par when we use the OIS as discount factor?

Does a bond paying floating coupon LIBOR, still has the value at par when we use the OIS as discount factor? It seems only when the Identity: $$B(t,T_2)(1+(T_2-T_1)F(t,T_1,T_2))=B(t,T_1)$$ still holds, the proposition above will be true. Here $B(t,T)$ is the value of zero coupon bond, $F(t,T_1,T_2)$ is the forward LIBOR.

In John Hull's book Options, Futures and Other Derivatives 9th page 205 ,shows the way to calculate the forward LIBOR implied in Swap rate under OIS discounting. But it's the case we know $B(t,T_1),$ but don't know $B(t,T_2).$

If we know both $B(t,T_1)$ and $B(t,T_2).$ Can we calculate the forward LIBOR still as above identity?

Denote

$D_{ois}(t):$ the discounted factor of OIS

$B(t,T):$ Bond price

$E_t[]:$ Conditional expectation at time $t$ under OIS-risk neutral measure which makes $D_{ois}(t)B(t,T)$ martingale for all $T.$

Use $N(t) = D_{ois}(t)B(t,T_1)$ as a numeraire to change the measure into OIS $T_1$-forward measure $E^{T_1}_t[]$(simply use expectation represent new measure).

Then $$\dfrac{B(t,T)}{B(t,T_1)} = \dfrac{D_{ois}(t)B(t,T)}{D_{ois}(t)B(t,T_1)}$$ should be martingale under $E^{T_1}_t[].$ Then use the definition of the forward LIBOR $F(t, T, T_{1})$ we can prove that $$\dfrac{1}{D_{ois}(T)}E_{T}\left[D_{ois}(T_1)\Big((T_1-T) \cdot F(T, T, T_{1})+1\Big)\right] = 1.$$

In a dual curve settings discounting is done at $B_{OIS}(t, T)$, whereas the forward libors are computed on the projection curve as $F(t, T_1, T_2) = (B_{libor}(t, T_1)/B_{libor}(t, T_2) - 1)/(T_2 - T_1)$, where $B_{OIS}(t, T)$ is the discount factor on the OIS curve and $B_{libor}(t, T)$ is the discount factor on the libor curve.
• actually I am not much understand what's the $B_{OIS}$ and $B_{LIBOR}?$ Why is discounted factor not D(t) and B(t,T)is the bond price. As you say, the forward LIBOR $F(t,T,T_1)$ will not change under the OIS discounting? – A.Oreo Oct 18 '17 at 16:01