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2

for an intuitive answer, if we start with a vanilla call as our base, then with an up & out call, we would like the underlying to go up in price yes. But as the price increases, we also increase the probability of kicking out and losing our payout - so we don't want it to go up too much. If the barrier is so far away that the probability of reaching it ...

2

Standard call options are trivially more expensive than up/down and out call options. However, for high strikes, down and out options will very likely never be knocked out, therefore their prices should be close to standard call options. For low strikes, down and out call options are almost worthless, therefore , the down and out call options curve price ...

2

The digital option pays $H$ at time $T$ if $S_T \geq K$ , so its option time at time $t$ is given by $$V_t=E_t\left[e^{-r(T-t)}H 1_{\{S_T \geq K\}}\right]=e^{-r(T-t)}H* P_t(S_T \geq K)$$ The model used is Black-model, that $$dS_t=rS_tdt+\sigma dW_t$$ or $$S_T=S_te^{\left(r-\frac12 \sigma^2\right)(T-t)+\sigma (W_T-W_t)}{}$$ Calculate $P_t(S_T \geq K)$ ...

2

Recall that the price of your contract is \begin{align*} V_t = e^{-r(T-t)} \mathbb{E}^\mathbb{Q} [H1_{\{S_T>K\}}|\mathcal{F}_t] \end{align*} because your option always pays $H$ if $S_T>K$. Next, \begin{align*} V_t &=He^{-r(T-t)} \mathbb{E}^\mathbb{Q} [1_{\{S_T>K\}}|\mathcal{F}_t] \\ &= He^{-r(T-t)} \mathbb{Q} [{\{S_T>K\}}|\mathcal{F}_t] \\...

1

$N\left(d_2\right)$ is the risk-neutral probability that the spot is greater than the strike at maturity, therefore the RN probability that you get your payoff.

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