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I am trying to better understand multi-curve bootstrapping, but I am clearly misunderstanding what is meant by:

a) projection curve

b) discount curve

I've tried googling the definitions but it's not making it any clearer.

Could someone please help give a definition and an example?

I'd thought that (for example) a 3m LIBOR curve would use a discounting curve (ie. Fed Funds) for tenors less that 3m and then the 3m LIBOR for tenors greater than 3m (projection curve).

But the more I read, the less this seems like a plausible definition.

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Let us examine what happens when we price our bread and butter, the vanilla interest rate swap in two worlds - the single curve world and the multi curve world.

Let the first reset date be $T_\alpha$ and the last payment date be $T_\beta$.

In the single curve world, the vanilla IRS has PV at time $t$ to be $$ \begin{align} \pi_t & = \mathbb{E}^{ \mathbb{Q} }_{t} \left[ \sum_{i} D_{tT_i} \tau_i \left[ L(T_{i-1};T_{i-1},T_i) - K \right] \right] \\ & = \sum_{i} P_{tT_i} \tau_i \left[ \mathbb{E}^{ \mathbb{Q}^{T_i} }_{t} \left[ L(T_{i-1};T_{i-1},T_i) \right] - K \right] \\ & = \sum_{i} P_{tT_i} \tau_i \left[ L(t;T_{i-1},T_i) - K \right] \\ & = \sum_{i} P_{tT_i} \tau_i L(t;T_{i-1},T_i) - K \sum_{i} P_{tT_i} \tau_i \\ & = \sum_{i} P_{tT_i} \tau_i \frac{1}{\tau_i} \left[ \frac{P_{tT_{i-1}} }{P_{tT_i}} -1 \right] - K \sum_{i} P_{tT_i} \tau_i \\ & = \sum_{i} P_{tT_i} \left[ \frac{P_{tT_{i-1}} }{P_{tT_i}} -1 \right] - K \sum_{i} P_{tT_i} \tau_i \\ & = P_{tT_\alpha} - P_{tT_\beta}-K \sum_{i} P_{tT_i} \tau_i \end{align} $$

In the multi curve world, the vanilla IRS has PV at time $t$ to be

$$ \begin{align} \pi_t & = \mathbb{E}^{ \mathbb{Q} }_{t} \left[ \sum_{i} D^{\text{ois}}_{tT_i} \tau^{\text{ois}}_i \left[ L(T_{i-1};T_{i-1},T_i) - K \right] \right] \\ & = \sum_{i} P^{\text{ois}}_{tT_i} \tau^{\text{ois}}_i \left[ \mathbb{E}^{ \mathbb{Q}^{T_i} }_{t} \left[ L(T_{i-1};T_{i-1},T_i) \right] - K \right] \\ & = \sum_{i} P^{\text{ois}}_{tT_i} \tau^{\text{ois}}_i \left[ L(t;T_{i-1},T_i) - K \right] \\ & = \sum_{i} P^{\text{ois}}_{tT_i} \tau^{\text{ois}}_i L(t;T_{i-1},T_i) - K \sum_{i} P^{\text{ois}}_{tT_i} \tau^{\text{ois}}_i \\ & = \sum_{i} P^{\text{ois}}_{tT_i} \tau^{\text{ois}}_i \frac{1}{\tau_i} \left[ \frac{P_{tT_{i-1}} }{P_{tT_i}} -1 \right] - K \sum_{i} P^{\text{ois}}_{tT_i} \tau^{\text{ois}}_i \\ \end{align} $$ Setting $\pi_t=0$, i.e., entering into the swap at time $t$ has no cost, means that the swap rate is $$ K=\frac{\sum_{i} P^{\text{ois}}_{tT_i} \tau^{\text{ois}}_i \frac{1}{\tau_i} \left[ \frac{P_{tT_{i-1}} }{P_{tT_i}} -1 \right] }{\sum_{i} P^{\text{ois}}_{tT_i} \tau^{\text{ois}}_i} $$

The difference is that now both ZCB curves are required to value the swap. The risk neutral measure $\mathbb{Q}$ is now explicitly under the discounting curve. You still assume that the projection curve is a martingale under $\mathbb{Q}$, though.

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If I promise to pay you 1000 USD year from now, we can use a Discounting Curve to find out how much this is worth in today's dollars. If I promise to pay you "3m LIBOR on a million dollars" a year from now, we need to do 2 steps: (1) Find out the market's current estimate of what 3m LIBOR will be, and convert it to dollars, (2) Discount this amount with the discount curve. The Projection Curve is used to perform Step (1).

The two curves are conceptually distinct, even if in the past the distinction was not considered important and the 2 curves were derived from the same underlying information (by using some simplifying assumptions). The Discount curve represents interest rates between now and a future date. The Projection curve refers to forward interest rates of 3mo tenor measured at a future date.

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  • $\begingroup$ The Discounting curve represents the cost of funds for the bank doing the discounting of any cash flow stream, the Projection curve refers to the specific rate (such as 3mo LIBOR) that underlies the particular swap we are valuing, Nowadays banks DO NOT finance themselves at 3mo LIBOR. That they do so was an (erroneous) simplifying assumption, now abandoned. $\endgroup$ – Alex C Nov 23 '19 at 18:55

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