# Swaptions Gamma Interview Questions

A while ago, I interviewed for a trader role and was given the below assignment (I didn't get the job). I wanted to revisit the questions to learn from my mistakes and be better prepared next time. Generally speaking, I have a good understanding of options and greeks but for some reason I'm having a hard time applying it here.

Question 1 Based on the P&L profile below please calculate a 25bps gamma profile for the up and down shocks.

I'm not sure what's meant by gamma profile or why the +/-50bp values are highlighted. I see that the changes in P&L are non-linear and asymmetrical, similar to the payoff diagram of a long put.

I know that

$$\\P\&L$$ = $$\delta$$ * $$\Delta$$S + $$\frac{1}{2}$$ * $$\Gamma$$ * ($$\Delta S^2$$)

but I'm not sure how to isolate the delta and gamma. Any pointers would be appreciated!

Question 2 The desk would like to purchase 5k of 25bps gamma with a goal of remaining delta and vega neutral. Utilizing the trades below develop the most cost efficient (cheapest 1m carry) way to achieve this goal.

To calculate 25bps Gamma, I used the following formula:

$$\Gamma_{25}$$ = $$\frac{Dn_{25} + Up_{25} - 2 * Base}{2 * Base * 0.0025^2}$$

which gives me the following values:

1m/10y = 102,204
3m/10y = 32,932
6m/10y = 16,095
1y/10y = 7,515
2y/10y = 3,671

I thought I should be able to re-calculate the Delta (10bp) as $$\\Up_{10} - Base$$ but the results don't make sense.

Can you confirm that my delta/gamma calculations are correct or tell me where I went wrong?

Finally, I tried to find a combination of straddles where the Gamma ~= 5k and Delta and Vega are close to 0 but I'm having a hard time finding a feasible solution, which makes me think my Gamma calculations might be incorrect.

• Hi there, could you point me to some elementary material on options & swap pricing, to better prepare for an interview like this? Oct 22 '18 at 1:54
Using Taylor polynomials of 2nd order:$$V(r+h)\approx V(r) + \frac{\partial{V}}{\partial{r}}h +\frac{1}{2}\frac{\partial^2{V}}{\partial{r}^2}h^2$$ $$V(r-h)\approx V(r) - \frac{\partial{V}}{\partial{r}}h +\frac{1}{2}\frac{\partial^2{V}}{\partial{r}^2}h^2$$
The sum of the previous 2 equation will give us gamma as: $$Gamma = \frac{\partial^2{V}}{\partial{r}^2} \approx \frac{V(r+h) -2V(r) + V(r-h)}{h^2}$$ whereas the difference of the two equations will give us delta as: $$Delta = \frac{\partial{V}}{\partial{r}} \approx \frac{V(r+h) -V(r-h)}{2h}$$
if you substitute Up10 for $$V(r+h)$$, Dn10 for $$V(r-h)$$ and 0.001 for $$h$$ in the Delta equation and multiply by 0.0001 (to get the 1 bp Delta) you will re-calculate the Delta (as presented in your table).
You can calculate Gamma in the same fashion (Up25 for $$V(r+h)$$, Dn25 for $$V(r-h)$$ and 0.0025 in gamma equation and multiply by 0.0001)