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Kevin
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Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$-measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e., \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

  • ThisThe above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable (due to the tower law).

  • If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A(S)-K,0\}$$\xi=\max\{A-K,0\}$ is allowed! It makes no differencesdifference whether you consider arithmetic or geometric averages here, or whether you use averages as strike prices. The sameRisk-neutral pricing also applies, for example, to other path-dependent exotic options such as (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That makes closed-form solutions very rare. Essentially, the distribution of the average $\int_t^T S_u\text{d}u$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a super-martingalesupermartingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...

Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$-measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e., \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

  • This above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable (due to the tower law).

  • If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A(S)-K,0\}$ is allowed! It makes no differences whether you consider arithmetic or geometric averages here, or whether you use averages as strike prices. The same applies, for example, to (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That makes closed-form solutions very rare. Essentially, the distribution of the average $\int_t^T S_u\text{d}u$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a super-martingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...

Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$-measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e., \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

  • The above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable (due to the tower law).

  • If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A-K,0\}$ is allowed! It makes no difference whether you consider arithmetic or geometric averages here, or whether you use averages as strike prices. Risk-neutral pricing also applies to other path-dependent exotic options such as (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That makes closed-form solutions very rare. Essentially, the distribution of the average $\int_t^T S_u\text{d}u$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a supermartingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...

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Kevin
Kevin
  • 16.4k
  • 4
  • 36
  • 68

Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$-measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e., \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

  • This above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable (due to the tower law).

  • If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A(S)-K,0\}$ is allowed! It makes no differences whether you consider arithmetic or geometric averages here, or whether you use averages as strike prices. The same applies, for example, to (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That makes closed-form solutions very rare. Essentially, the distribution of the average $\int_t^T S_t\text{d}t$$\int_t^T S_u\text{d}u$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a super-martingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...

Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$-measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e., \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

  • This above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable (due to the tower law).

  • If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A(S)-K,0\}$ is allowed! It makes no differences whether you consider arithmetic or geometric averages here, or whether you use averages as strike prices. The same applies, for example, to (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That makes closed-form solutions very rare. Essentially, the distribution of the average $\int_t^T S_t\text{d}t$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a super-martingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...

Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$-measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e., \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

  • This above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable (due to the tower law).

  • If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A(S)-K,0\}$ is allowed! It makes no differences whether you consider arithmetic or geometric averages here, or whether you use averages as strike prices. The same applies, for example, to (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That makes closed-form solutions very rare. Essentially, the distribution of the average $\int_t^T S_u\text{d}u$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a super-martingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...

added 37 characters in body
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Kevin
Kevin
  • 16.4k
  • 4
  • 36
  • 68

Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$ measurable-measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e., \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

  • This above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable (due to the tower law).

  • If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A(S)-K,0\}$ is allowed! It makes no differences whether you haveconsider arithmetic or geometric averageaverages here, or whether you use averages as strike prices. The same applies, for example, to (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity results for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That'sThat makes closed-form solutions are very rare. Essentially, the distribution of the average $\int_t^T S_t\text{d}t$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a super-martingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...

Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$ measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e. \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

  • This above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable.

  • If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A(S)-K,0\}$ is allowed! It makes no differences whether you have arithmetic or geometric average here, or whether you use averages as strike prices. The same applies to (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity results for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That's closed-form solutions are very rare. Essentially, the distribution of the average $\int_t^T S_t\text{d}t$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a super-martingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...

Risk-neutral pricing

A time-$T$ payoff is an integrable, $\mathcal{F}_T$-measurable random variable $\xi$. The value process of the discounted payoff is then a $\mathbb{Q}$-martingale, i.e., \begin{align*} V_t=\mathbb{E}^\mathbb{Q}_t\left[\frac{B_t}{B_T}\xi\right], \end{align*} where $B_t$ is a locally risk-free bank account ($\text{d}B_t=r_tB_t\text{d}t$).

  • This above result essentially follows from the definition of $\mathbb{Q}$ and the fact that $M_t=\mathbb{E}_t[X]$ is a martingale if $X$ is integrable (due to the tower law).

  • If $r_t\equiv r$ is constant, we have $B_t=e^{rt}$ and $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\xi\right]$.

Does it work? Yes!

The only requirement is that $\xi$ is known (observable, measurable) at maturity $T$. There is no requirement that $\xi$ needs to be path-independent. Thus, $\xi$ can indeed be an average and standard risk-neutral pricing applies to (European-style) Asian options, i.e., $\xi=\max\{A(S)-K,0\}$ is allowed! It makes no differences whether you consider arithmetic or geometric averages here, or whether you use averages as strike prices. The same applies, for example, to (European-style) barrier options.

Indeed, semi-closed-form numerical methods for Asian options rely on explicitly on this risk-neutral pricing framework.

We also get some simple results: The identity $\max\{x-K,0\}-\max\{K-x,0\}=x-K$ gives a put-call parity for Asian options.

Where's the problem?

The only problem is that computing the first moment of the option payoff is darn difficult. Most often, we're interested in arithmetic Asian options but we tend to model stock prices in an exponential form. That makes closed-form solutions very rare. Essentially, the distribution of the average $\int_t^T S_t\text{d}t$ is not really known for sensible stock price models. For geometric averages, the situation is a bit better.

American options

The risk-neutral pricing formula does not apply to early exercise features (e.g., American put options). Their prices relate to the Snell envelope, which is a super-martingale, see this answer. Their prices can thus be decomposed into a European option (a martingale) and an early exercise correction term (Riesz decomposition or Doob-Meyer decomposition). The maths for these early exercise features is more difficult. Obviously, pricing American-style Asian options is a really difficult task (I'd opt for MC simulations)...

Source Link
Kevin
Kevin
  • 16.4k
  • 4
  • 36
  • 68