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I have a task to do but it is very difficult.. I have to calculate the:

  1. Delta Sensitivity analytically, that is the first derivative of caplet price wrt the forward rate, using the black model to price a generic caplet
  2. Delta Vega analytically, that is the first derivative of caplet price wrt the volatility, using the black model to price a generic caplet

How can I obtain that?

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I'm not sure I agree with that being a very difficult task...

The black formula for a caplet (using notation from Hull's book) is given by:

$caplet = L \delta_k P(0, t_{k+1}) [F_k N(d_1) - R_kN(d_2)]$

where:

$d_1 = \frac{ln(F_k/R_k) + \sigma_k^2t_k/2}{\sigma_k\sqrt{t_k}}$ and $d_2 = d_1 - \sigma_K \sqrt{t_k}$

The delta will just be the first derivative of the equation above in respect to F, ie:

Delta = $ L \delta_k P(0, t_{k+1}) N(d_1)$

You can test this by comparing the calculated Delta * 1 basis point on one side with the caplet price for forward + 1 basis point minus the caplet price for forward - 1 basis point divided by two.

For the vega, just do the same thing, but derive in respect to $\sigma$. (hint: the sigma is in the $d_1$)

Edit after comments:

As Jan Stuller very well pointed out, the derivation of the well known result for the delta is, in fact, not as simple as it seems, although not that difficult.

Ignoring the constants $L \delta_k P(0, t_{k+1})$ ...

$$ \Delta = \frac{\partial c}{\partial F} = N(d_1) + F \frac{\partial N(d_1)}{\partial F} - R_k \frac{\partial N(d_2)}{\partial F}$$

Next we apply the chain rule:

$$ \frac{\partial N(d_1)}{\partial F} = \frac{\partial N(d_1)}{\partial d_1} \frac{\partial d_1}{\partial F}$$

and so:

$$ \Delta = N(d_1) + F \frac{\partial N(d_1)}{\partial d_1} \frac{\partial d_1}{\partial F} - R_k \frac{\partial N(d_2)}{\partial d_2} \frac{\partial d_2}{\partial F}$$

First thing to notice is that since:

$d_1 = \frac{ln(F_k/R_k) + \sigma_k^2t_k/2}{\sigma_k\sqrt{t_k}}$ and $d_2 = d_1 - \sigma_K \sqrt{t_k}$

we have:

$$\frac{\partial d_1}{\partial F} = \frac{\partial d_2}{\partial F} = \frac{1}{F_k \sigma \sqrt{t_k}}$$

and so:

$$ \Delta = N(d_1) + \frac{1}{F_k \sigma \sqrt{t_k}} \left[ F \frac{\partial N(d_1)}{\partial d_1} - R_k \frac{\partial N(d_2)}{\partial d_2} \right] $$

Second, lets see how we can simplify:

$$ \left[ F \frac{\partial N(d_1)}{\partial d_1} - R_k \frac{\partial N(d_2)}{\partial d_2} \right]$$

Considering that:

$$N(d_1) = \frac{1}{\sqrt{2\pi}} \int^{d_1}_{-\infty} e^{-x^2 / 2} dx$$

it follows that:

$$\frac{\partial N(d_1)}{\partial d_1} = N^{'}(d_1) = \frac{1}{\sqrt{2\pi}} e^{-d_1^2 / 2}$$

and so:

$$ \left[ F \frac{\partial N(d_1)}{\partial d_1} - R_k \frac{\partial N(d_2)}{\partial d_2} \right] = \left[ FN^{'}(d_1) - R_k N^{'}(d_2) \right]$$

Now with a simple substitution of $d_2$ for $d_1 - \sigma_k \sqrt{t_k}$, we have:

$$R_k N^{'}(d_2) = R_k N^{'}(d_1 - \sigma_k \sqrt{t_k})$$

$$ = R_k \frac{1}{\sqrt{2\pi}} e^{\frac{-(d_1 - \sigma_k \sqrt{t_k})^2}{2}} = R_k \frac{1}{\sqrt{2\pi}} e^{\frac{-d_1^2}{2}} e^{ \frac{-(2d_1 \sigma_k \sqrt{t_k} + \sigma_t^2 t_k)}{2}}$$

$$ = R_k N^{'}(d_1) e^{ \frac{ln(F_k/R_k) + \sigma_k^2t_k/2}{\sigma_k\sqrt{t_k}} \sigma_k \sqrt{t_k}} e^{- \sigma_t^2 t_k / 2)} = R_k N^{'}(d_1) e^{ ln(F_k/R_k) + \sigma_k^2t_k/2} e^{- \sigma_t^2 t_k / 2)}$$

$$ = R_k N^{'}(d_1) e^{ ln(F_k/R_k) + \sigma_k^2t_k/2 - \sigma_t^2 t_k / 2)} = R_k N^{'}(d_1) \frac{F_k}{R_k} = F_k N^{'}(d_1)$$

$$ $$

and so

$$R_k N^{'}(d_2) = F_k N^{'}(d_1)$$

$$F_k N^{'}(d_1) - R_k N^{'}(d_2) = 0$$

So finally, the delta of the caplet would be:

$$ \Delta = N(d_1) + \frac{1}{F_k \sigma \sqrt{t_k}} \left[ FN^{'}(d_1) - R_k N^{'}(d_2) \right] = N(d_1) + \frac{1}{F_k \sigma \sqrt{t_k}} \times 0 $$

$$ \Delta = N(d_1)$$

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    $\begingroup$ Good answer, David. Just wanted to point out that when taking the derivative w.r.t. $F$, one also has to worry about the $F$ being in the $d_1$ and $d_2$ terms. The way the math works is that the terms coming out of $$\frac{\partial N(d_1)}{\partial d_1}\frac{\partial d_1}{\partial F}$$ & $$\frac{\partial N(d_2)}{\partial d_2}\frac{\partial d_2}{\partial F}$$ happen to cancel each other. So it then looks as if one can just differentiate $caplet$ w.r.t. $F$ and "pretend" that $F$ isn't inside $d_1$ & $d_2$, but that's just a "coincidence". I know you know: but maybe the OP doesn't know...:) $\endgroup$ Jan 15, 2021 at 16:41
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    $\begingroup$ PS: Obviously same with the classical Black-Scholes. I have seen far too often even experienced partitioners "thinking" that BS delta is $N(d_1)$ because $\frac{\partial}{\partial S}$ of a call is $N(d_1)$ "of course, because $\frac{\partial S}{\partial S}$ equals to 1 "... $\endgroup$ Jan 15, 2021 at 16:44

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