When an Arithmetic Asian Call Can Exceed a Vanilla Call’s Payoff

# Asian vs Vanilla Call


# Asian vs Vanilla Call












When would an asian call be more expensive than a vanilla call, all things being equal ? Assume fixed strike, arithmetic average.

I thought that it was mainly about forward and volatility term structure.

For example, I thought that it could be the case if we had : average of forward prices between 0 and T > forward price at time T, assuming a flat volatility term structure

Don't know how I could combine both forwards and term structures exactly to explain this, or even if my assumptions are right.

## Answer by blenton (score -1)

https://quant.stackexchange.com/a/82463

The price for each option is just the present value of the payout. We assume the maturity is the same, so the payouts of both options are discounted by the same factor. And since payout is strictly non-negative, the one with the higher payout is more expensive.

Let P denote payout, and upon maturity time T $$P_{vanilla} (T) = max(0, S(T) - K)$$ $$P_{asian} (T) = max(0, \frac{1}{T}\sum^T_{t=1}S(t) - K)$$ So, Asian call will be more expensive if $\frac{1}{T}\sum^T_{t=1}S(t) \gt S(T)$ and $max(0, \frac{1}{T}\sum^T_{t=1}S(t) - K) > 0$. See the example below

```
import numpy as np
import matplotlib.pyplot as plt

def vanilla_payout(K, S):
  return max(0, S[-1] - K)
def asian_payout(K, S):
  return max(0, S.mean() - K)
K = 95
T = 255
S = np.array(np.linspace(100,96,T)+np.random.random(T))
print("Vanilla Payout", vanilla_payout(K, S))
print("Asian Payout", asian_payout(K, S))
fig, ax = plt.subplots()
ax.plot(S)
ax.scatter(T, S[-1], color='red', label="Spot(T)")
ax.hlines(S.mean(), 0, T, linestyles='dashed', color='red', label="Average(T)")
ax.hlines(K, 0, T, linestyles='dashed', color='orange', label="strike")
plt.legend()
plt.show()
```