Expected Put Payoff at an Independent Exponential Exercise Time

# American put option. Exercise time is a random variable, calculation of expected payoff


# American put option. Exercise time is a random variable, calculation of expected payoff












I got an American put option, where the payoff is $V_\tau = \max(K - X_{\tau}, 0)$ and $X_{\tau}$ is the price of an underlying at the stopping time $\tau < T$. The underlying follows a standard GBM with $r = q = 0$; $X_0$ is given.

I need to calculate the expectation $E[V]$ under the assumption that $\tau$ has exponential distribution with intensity $\lambda = 0.025$.

I tried transforming this equation into: $$\int_0^\infty (K - X_0e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau}Z})^+\lambda e^{-\lambda \tau}d\tau$$ but then I'm just completely lost with how to proceed with the square root. I know that by definition $E[\tau] = \frac{1}{\lambda}$ but can I use this as an answer? As in, can I claim that: $$E[V] = V\left(X_{\frac{1}{\lambda}}, \frac{1}{\lambda}\right) \text{ ?}$$

## Answer by carry_and_pray (score 1)

https://quant.stackexchange.com/a/85605

In general you cannot replace the random time by its mean and write, $$ \mathbb{E} [(K - X_\tau)^+] = (K - X_{1/\lambda})^+ \quad\text{or}\quad \mathbb{E}[(K - X_\tau)^+] = P_\text{put} (X_0, 1 / \lambda). $$ The payoff is non-linear in both $X$ and $t$ so you should average over the full distribution of $\tau$ and not just its mean. Also, once the exercise time $\tau$ is given exogenously and is independent of the stock, this is not really an American option problem anymore and is just a put payoff evaluated at a random time.

Suppose $dX_t = \sigma X_t dW_t$ with $X_0 = x$ since $r = q = 0$. Then $X_t = x \exp \left( - \frac{1}{2} \sigma^2 t + \sigma W_t \right)$. I'll also first assume that $\tau \sim \text{Exp}(\lambda)$ and $P(\tau \in dt) = \lambda e^{-\lambda t} dt$ and that $\tau$ is independent of $W$.

Then by conditioning on $\tau = t$, $$ \mathbb{E}[(K - X_\tau)^+] = \int_0^\infty \lambda e^{-\lambda t} \mathbb{E}[(K - X_t)^+] dt. $$ However, for fixed $t$, we know that $\mathbb{E}[(K - X_t)^+]$ is just the Black-Scholes European put price with maturity $t$, zero rates and spot $x$ so, $$ \mathbb{E}[(K - X_t)^+] = K \Phi(-d_2(t)) - x \Phi(-d_1(t)), $$ so we already have what $\mathbb{E}[(K - X_\tau)^+]$ which is, $$ \mathbb{E}[(K - X_\tau)^+] = \int_0 \lambda e^{-\lambda t} \left(K \Phi(-d_2(t)) - x \Phi(-d_1(t)) \right) dt. $$

In this exponential-time case, you could actually go further to get a closed form. I'll just define $u(x) := \mathbb{E}_x [(K - X_\tau)^+]$ and using the exponential density $u(x) = \mathbb{E}_x \left[ \int_0^\infty \lambda e^{-\lambda t} (K - X_t)^+ dt \right]$.

Since the generator of $X$ is $L f(x) = \frac{1}{2} \sigma^2 x^2 f''(x)$, the resolvent equation is just $\lambda u(x) - L u(x) = \lambda (K - x)^+$ that is, $$ \lambda u(x) - \frac{1}{2} \sigma^2 x^2 u''(x) = \lambda (K - x)^+. $$ Now we can just solve this separately on $x < K$ and $x > K$.

For $x > K$, the rhs is 0 so $\frac{1}{2} \sigma^2 x^2 u'' - \lambda u = 0$. Try $u(x) = x^m$ so then $\frac{1}{2} \sigma^2 m(m - 1) - \lambda = 0$.

The roots are $m - \alpha, \beta$ where $\alpha = \frac{1 + \nu}{2}$ and $\beta = \frac{1 - \nu}{2}$ and $\nu := \sqrt{1 + \frac{8\lambda}{\sigma^2}}$.

Hence $u(x) = Ax^\alpha + B x^\beta$ for $x > K$. Since $\alpha > 1$ and $\beta < 0$, boundedness as $x \to \infty$ forces $A = 0$ so $u(x) = Cx^\beta$ for $x > K$.

For $x < K$ the equation is $\frac{1}{2} \sigma^2 x^2 u'' - \lambda u = -\lambda (K - x)$ so a plain particular solution is $u_p(x) = K - x$ because $u_p '' = 0$. So $u(x) = K - x + A x^\alpha + \beta x^\beta$ for $x < K$. Boundedness as $x \to 0$ forces $B = 0$ since $\beta < 0$. Thus $u(x) = K - x + Ax^\alpha$ for $x < K$.

Now matching $u$ and $u'$ continuously as $x = K$ gives $AK^\alpha = CK^\beta$. Differentiating gives $-1 + A\alpha K^{\alpha - 1} = C \beta K^{\beta - 1}$. Using $C = AK^{\alpha - \beta}$ makes, $$ -1 + A(\alpha - \beta) K^{\alpha - 1} = 0. $$ Since $\alpha - \beta = \nu$ then $A = \frac{K^{1 - \alpha}}{\nu}$ and $C = \frac{K^{1 - \beta}}{\nu}$.

Therefore the final closed form is, $$ u(x) = \begin{cases} K - x + \frac{K}{\nu} \left( \frac{x}{K} \right)^\alpha, &\quad 0 < x \le K,\\ \frac{K}{\nu} \left( \frac{x}{K} \right)^\beta, &\quad x \ge K \end{cases}. $$

Also, you state that $\tau < T$ but an exponential random variable is not automatically bounded by $T$. If what you really mean is exercise at $\tau \wedge T = \min(\tau, T)$ then the expectation becomes, $$ \mathbb{E}[(K - X_{\tau \wedge T})^+] = \int_0^T \lambda e^{-\lambda t} \mathbb{E}[(K - X_t)^+] dt + e^{-\lambda T} \mathbb{E}[(K - X_T)^+], $$ which is just, $$ \mathbb{E}[(K - X_{\tau \wedge T})^+] = \int_0^T \lambda e^{-\lambda t} P_\text{BS}^\text{put}(x, K, t) dt + e^{-\lambda T} P_\text{BS}^\text{put} (x, K, T). $$

## Answer by Charles Fox (score 0)

https://quant.stackexchange.com/a/45132

You cannot make that claim. $v(T,s_0,k...)$ increases with approximately of $\sqrt t$. It is not linear with respect to $t$. By Jenson's inequality $E[v(T,s_0,k...)] < v(E[T],s_0,k...)$ when $v''(T)<0$ and $T$ is not a constant.

## Answer by Valometrics.com (score 0)

https://quant.stackexchange.com/a/50679

The best way to price this kind of options is to use monte carlo: 1. You generate the stopping time. 2. you generate the underlying value at this generated stopping time and you store it in a vector. 3. The price will be the mean of your vector as r and q are equal to zero. That's all.

## Answer by kwinto (score 0)

https://quant.stackexchange.com/a/74826

Let's recall why no analytical formula is known for the price of American Put option. This is due to the early exercise condition, which requires that options's price should be bigger or equal to its exercise payoff for any $S_t$ at any time $t \le T$ (in other words $V(S_t,t) \ge max(K-S_t, 0)$). This property is also known as path-dependence, as payoff at maturity $T$ doesn't depend only on $S_T$, but also on previous values of $S_t$ where it could be potentially exercised. An arbitrage opportunity exist when this condition is broken: exercise put and instantly buy it back with a net profit OR buy put and instantly exercise.

Early exercise condition is very hard to account for in analytical solution, hence numerical methods are used. Most popular are Finite Difference and Monte-Carlo. In both cases you explicitly ensure that put price doesn't fall below the exercise payoff. (Finite difference works well for 1D and 2D problems and is easier to implement, however Monte-Carlo is more stable in higher dimensions but is harder to code, again due to the early exercise condition.)

In your problem, a new exercise condition is introduced, which is not path-dependent, but random (distributed exponentially). It might be interpreted as

- Replacement for the path-dependent condition. In this case exercise can be initiated only by some random chance and holder has no right to exercise. As path-dependent condition was lifted, we have a chance for a closed-form solution (assuming $r=q=0$): $$ E[V] = \int_0^T d\tau \ \lambda e^{-\lambda \tau} \int_0^{K} ds \ (K-S) \ \rho_s(S,\tau) $$ Here $\rho_s(s,t)$ is distribution density of $S$ at time $t$ in risk-neutral measure: $$ \rho_s(s,t) = \frac {s_0} {\sqrt{2\pi} \ s\ \sigma\sqrt{t}} \exp\left( - \frac{\left(\ln s / s_0 + 0.5 \sigma^2 t \right)^2}{2\sigma^2 t}\right) $$ The last integral is BS price of European Put with maturity $\tau$, hence you need to find $$ E[V] = \int_0^T dt \ \lambda e^{-\lambda t} \ V_{put}(S_0, t) $$

- Addon to the path-dependent condition. In this case exercise can be initiated by some random chance or by the holder. This is a "loosing" strategy as by randomly exercising we always lose extrinsic value (which is non-negative for American options). This is more challenging as path-dependence is still present. Not sure if you can incorporate exercise randomness into finite-difference approach, so what is left is Monte-Carlo. It will be 2D, as you need to simulate both $S_t$ and exercise time $\tau$ (if exercise happens). Otherwise. it should not be much different from standard American pricing a la Longstaff-Schwartz.

Hopefully, these thoughts will be helpful.

## Answer by Johnny Iwash (score 0)

https://quant.stackexchange.com/a/80141

I haven't seen an answer that answers how I would answer, this is what I think you are asking, you have a stopping time $\min(\tau,T)$ where $\tau$ is exponentially distributed with rate $\lambda$, so you can condition on the disjoint sets $\tau \leq T$ and $\tau > T$, then you have a GBM independent of the stopping time which is

$$ dX_t = \sigma X_t dW_t, \quad X_0 = x>0 $$ which admits a closed-form solution $$ X_t = X_0 e^{\sigma W_t} $$

and you want to compute \begin{align} \mathbb{E}(\max(K-X_{\min(\tau,T)},0)=&\mathbb{P}(\tau < T)\mathbb{E}(\max(K-X_{\min(\tau,T)},0 )\mid \tau < T)\\+&\mathbb{P}(\tau \geq T)\mathbb{E}(\max(K-X_{\min(\tau,T)},0)\mid \tau \geq T) \end{align} $\mathbb{P}(\tau \geq T)=e^{-\lambda T}$ so you got here a closed form which is \begin{align} \mathbb{E}(\max(K-X_{\min(\tau,T)},0)=&\int_{-\infty}^{\infty}\int_0^T dx d\tau \lambda e^{-\tau \lambda} p_{\tau \sigma^2}(x)\max(V-x,0) \\ +e^{-\lambda T}&\int_{-\infty}^\infty dx p_{\sigma^2 T}(x) \max(V-x,0) \end{align} where $p_{\sigma^2}(x)$ is the density function for $\mathcal{N}(0,\sigma^2)$, you can condition again over $e^{\sigma W_T} \geq V$ in the second term to get a "closer-form" (haha) and I guess in the first one too using barrier arguments, maybe there are martingale arguments also because notice how your stopping time is markovian. The integral I wrote should probably be done numerically, it is just in $\mathbb{R}\times [0,T]$.

## Answer by Hans (score -1)

https://quant.stackexchange.com/a/54829

Substitute $x=\sqrt{\tau}$. There may be two intervals of $[0,\infty)$ for the integral over $\tau$ that is effective under the function $(\cdot)^+$. For convenience, we take the right most interval starting from $x^2_1$. \begin{align} &\frac12\int_{x^2_1}^\infty (K - X_0e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau}Z}) e^{-\lambda \tau}d\tau \\ =& \int_{x_1}^\infty (K - X_0e^{-\frac{1}{2}\sigma^2 x^2 + \sigma xZ}) e^{-\lambda x^2}xdx \\ =& \frac{K}{2\lambda}e^{-\lambda x_1^2}-X_0\Big(\int_{x_1}^\infty e^{-\frac12a(x -b)^2 +c}(x-b) dx +b\int_{x_1}^\infty e^{-\frac12a(x -b)^2 +c}dx \Big) \\ =& \frac{K}{2\lambda}e^{-\lambda x_1^2}-X_0 \Big(\frac12 e^{-\frac12a(x_1 -b)^2 +c}+bF(x_1)\Big) \end{align} where $a, b, c$ are appropriate constants and $F(x)$ is essentially the complementary error function.