<p>Generally:</p>

<p>$P(f\in[10,20]) = P(120 \leq S_t \leq 130) = P(S_t \leq 130) - P(S_t \leq 120)$ 
That is, the probability that the option is between 10 and twenty is the same that the stock is between 120 and 130. The probability that the stock is between 120 and 130 is the probability the stock is less than 130 minus the probability that it is less than 120. If you are a more visual person consider this picture:
<img src="https://i.sstatic.net/aRqX5.gif" alt="CDF to calculate interior prob"> </p>

<p><a href="http://www.mathwave.com/articles/calculate-probability-distributions.html" rel="nofollow noreferrer">Source MathWave</a></p>

<p>If we know that $W_t$ follows a driftless-Wiener process that:</p>

<p>$W_t - W_0 \sim N(0, 15^2t) \rightarrow S_t \sim N(95, 15^2t)$</p>

<p>If $S_t \sim N(95, 15^2t)$ then $P(S_t \leq 130) = $ prob. that a normal random variable with mean 95 and std. deviation 15t which is just the cumulative distribution function (CDF) of that normal random variable. </p>

<p>Here, that CDF is $\Phi( (x-95) / (15 \sqrt(t)))$, where $\Phi$ is the CDF of the standard normal. So the answer is $$\Phi( (130-95) / (15 \sqrt t)) - \Phi( (120-95) / (15 \sqrt t))$$</p>

<p>Update:
I thought that this was a simple option payoff but this is actually a trickier problem because it is a look back option taking the maximum stock value from 0 to 2. The general idea with this more complex problem is we are studying the maximum of a Wiener process motion rather than the Wiener process itself. <a href="http://www.yats.com/doc/stats-en.pdf" rel="nofollow noreferrer">Stochastic Processes and Related Distributions by Daniel Herlemont</a>, section 2 should be of some help. But I don't have a solution to this problem at this time. </p>

CC BY-SA 3.0
2015-02-10T16:31:35.213Z
Thanks for image. But where we use the fact that $T=2$?
luka5z
CC BY-SA 3.0
2015-02-10T16:54:15.607Z
Just plug in 2 for $t$ in the final equation. The solution is for all $t$.
BKay
CC BY-SA 3.0
2015-02-10T17:00:44.940Z
The last thing - why we just left $\max $ operator?
luka5z
CC BY-SA 3.0
2015-02-10T17:14:19.423Z
Oh boy, I misread the question. I thought it was max of  st - 110 and 0, a conventional option. This is a look back option and I'm not sure I can easily present a solution to the look back option. You should take away my "answered" status.
BKay
CC BY-SA 3.0
2015-02-10T17:48:19.043Z
no problem, it happens
luka5z
CC BY-SA 3.0
2015-02-10T16:20:27.943Z
2015-02-10T18:38:18.230Z
2015-02-10T18:38:18.230Z
BKay
<p>Maybe I'm familiar with a completely different background of mathematical finance, but I'll drop my two cents anyways.</p>

<p>First, note that </p>

<p>$$ \int d Wt $$ </p>

<p>is a stochastic integral and not easily integrable with standard methods. I hope you are preset with a tool of stochastic calculus.</p>

<p>The following method based on no-arbitrage was first introduced by Merton.</p>

<ul>
<li>We look for a derivative price based on the price of $S_t$, denoted by $\Pi(S_0)$</li>
<li>You form a portfolio based on the underlying $S$ and the derivative $f$, with portfolio dynamics $dV_t = V_t (\omega_t^s \frac{dS_t}{S_t} + \omega_t^f \frac{df}{f} )$, where $\omega$ denotes the two weights on the portfolio</li>
<li>These weights are chosen such that we "kill" the stochastic component $dW$</li>
<li>We now have $dV_t = V_t k dt$</li>
<li>Due to no-arbitrage, $k$ must be zero (there is no growth component on $S_t$ either)</li>
<li>solve for $f$</li>
</ul>

CC BY-SA 3.0
2015-02-09T21:40:51.390Z
2015-02-09T21:40:51.390Z
FooBar
<p>In moment $t=0$ we bought option with expiration date $T=2$. The payoff function of this option is given by:</p>

<p>$$f=(\max_{t\in[0,T]} S_t -110)^{+}$$</p>

<p>where $S_t$ satisfies</p>

<p>$$dS_t=15dW_t$$
$$S_0=95$$</p>

<p>Find probability $P(f\in[10,20])$. Use cumulative density function of standard normal distribution to express it.</p>

<hr>

<p>Obviously</p>

<p>$$S_t=95+15W_t$$</p>

<p>But i don't have any idea how can I move further. Any help please?</p>

<stochastic-processes><stochastic-calculus><asset-pricing>
Find probability that payoff function is in $[10,20]$
luka5z
# Find probability that payoff function is in $[10,20]$

(Asked by: username_0 on 2015-02-09)

In moment $t=0$ we bought option with expiration date $T=2$. The payoff function of this option is given by:

$$f=(\max\_{t\in[0,T]} S\_t -110)^{+}$$

where $S\_t$ satisfies

$$dS\_t=15dW\_t$$
$$S\_0=95$$

Find probability $P(f\in[10,20])$. Use cumulative density function of standard normal distribution to express it.

---

Obviously

$$S\_t=95+15W\_t$$

But i don't have any idea how can I move further. Any help please?

**Tags:** stochastic-processes, stochastic-calculus, asset-pricing


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## 2 answers

### Answer by username_1 on 2015-02-10. Score: 3

Generally:

$P(f\in[10,20]) = P(120 \leq S\_t \leq 130) = P(S\_t \leq 130) - P(S\_t \leq 120)$
That is, the probability that the option is between 10 and twenty is the same that the stock is between 120 and 130. The probability that the stock is between 120 and 130 is the probability the stock is less than 130 minus the probability that it is less than 120. If you are a more visual person consider this picture:
![CDF to calculate interior prob](https://i.sstatic.net/aRqX5.gif)

[Source MathWave](http://www.mathwave.com/articles/calculate-probability-distributions.html)

If we know that $W\_t$ follows a driftless-Wiener process that:

$W\_t - W\_0 \sim N(0, 15^2t) \rightarrow S\_t \sim N(95, 15^2t)$

If $S\_t \sim N(95, 15^2t)$ then $P(S\_t \leq 130) = $ prob. that a normal random variable with mean 95 and std. deviation 15t which is just the cumulative distribution function (CDF) of that normal random variable.

Here, that CDF is $\Phi( (x-95) / (15 \sqrt(t)))$, where $\Phi$ is the CDF of the standard normal. So the answer is $$\Phi( (130-95) / (15 \sqrt t)) - \Phi( (120-95) / (15 \sqrt t))$$

Update:
I thought that this was a simple option payoff but this is actually a trickier problem because it is a look back option taking the maximum stock value from 0 to 2. The general idea with this more complex problem is we are studying the maximum of a Wiener process motion rather than the Wiener process itself. [Stochastic Processes and Related Distributions by Daniel Herlemont](http://www.yats.com/doc/stats-en.pdf), section 2 should be of some help. But I don't have a solution to this problem at this time.

> **Comments:**
> **username_0**: Thanks for image. But where we use the fact that $T=2$?  
> **username_1**: Just plug in 2 for $t$ in the final equation. The solution is for all $t$.  
> **username_0**: The last thing - why we just left $\max $ operator?  
> **username_1**: Oh boy, I misread the question. I thought it was max of  st - 110 and 0, a conventional option. This is a look back option and I'm not sure I can easily present a solution to the look back option. You should take away my "answered" status.  
> **username_0**: no problem, it happens  

---

### Answer by username_2 on 2015-02-09. Score: 2

Maybe I'm familiar with a completely different background of mathematical finance, but I'll drop my two cents anyways.

First, note that

$$ \int d Wt $$

is a stochastic integral and not easily integrable with standard methods. I hope you are preset with a tool of stochastic calculus.

The following method based on no-arbitrage was first introduced by Merton.

* We look for a derivative price based on the price of $S\_t$, denoted by $\Pi(S\_0)$
* You form a portfolio based on the underlying $S$ and the derivative $f$, with portfolio dynamics $dV\_t = V\_t (\omega\_t^s \frac{dS\_t}{S\_t} + \omega\_t^f \frac{df}{f} )$, where $\omega$ denotes the two weights on the portfolio
* These weights are chosen such that we "kill" the stochastic component $dW$
* We now have $dV\_t = V\_t k dt$
* Due to no-arbitrage, $k$ must be zero (there is no growth component on $S\_t$ either)
* solve for $f$


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