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I am looking into pricing options in finance and I came across the following problem: I want to simplify the probability:

$\mathbf{Q}(z_1 > - \hat{d}_1, z_2 > \hat{d}_2), $

where $z_1$ and $z_2$ are two standard normal variables such that $z_1 = \frac{1}{\sqrt{T_0}}W^{*}_{T_0}$ and $z_2 = \frac{1}{\sqrt{T}}W^{*}_{T}$, where $T>T_0$ and $W^{*}_{T}$ is the standard Brownian motion on the increment $T$. Furthermore, $\hat{d}_1$ and $\hat{d}_2$ are described as:

$\hat{d}_1 = \frac{\text{ln}\big(\frac{S(0)}{\bar{x}}\big) + (r - \frac{1}{2}\sigma^2)T_0}{\sigma\sqrt{T_0}},$

$\hat{d}_2 = \frac{\text{ln}\big(\frac{S(0)}{K}\big) + (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}.$

where $S(0)$ is the stock price at time $0$, $r$ is the risk-free rate, $\sigma^2$ is the annualized volatility and $K$ and $\bar{x}$ are constants.

I am quite sure that the distribution I am looking for is the bivariate normal, but I have no idea how to get there.

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1 Answer 1

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We first note that we can work directly with the underlying Brownian motion, rather than the scaled variables $z_1, \,z_2$ since

$$\mathbf Q \left( z_1 > \hat d_1, \, z_2 > \hat d_2\right) = \mathbf Q \left( W_{T_0} > \sqrt{T_0}\, \hat d_1, \, W_{T} > \sqrt{T}\, \hat d_2\right).$$

Henceforth I denote $d_1= \sqrt{T_0} \, \hat d_1, \, d_2 = \sqrt{T}\, \hat d_2$.

To compute the probability, we first use the law of total probability conditioning on the event $\{W_{T_0} \in dx\}$, and then use independence of Brownian increments. That is

\begin{align*} \mathbf Q \left( W_{T_0} > d_1, W_T > d_2\right) & = \int_{\mathbf R} \mathbf Q \left(W_{T_0} > d_1,\, W_T > d_2 \, | \, W_{T_0} \in dx\right)\, \mathbf Q\left( W_{T_0} \in dx\right)dx \\ & = \int_{d_1}^\infty \mathbf Q \left( W_T > d_2 \, | \, W_{T_0} \in dx\right)\, \mathbf Q\left( W_{T_0} \in dx\right)dx \\ & = \int_{d_1}^\infty \mathbf Q \left( W_T - W_{T_0} >d_2 -x\right)\, \mathbf Q\left( W_{T_0} \in dx\right)dx \\ \end{align*} Then we can use that $$W_{T_0} \sim N(0, T_0), \qquad W_{T} - W_{T_0} \sim N(0, T - T_0)$$ are independent and Normally distributed; let $\Phi$ denote the CDF of a standard normal distribution, then from the above we have

\begin{align*} \mathbf Q \left( W_{T_0} > d_1, W_T > d_2\right) & = \int_{d_1}^\infty \left(1 - \Phi\left( \frac{d_2 - x}{\sqrt{T - T_0} } \right) \right) \frac{1}{\sqrt{T_0}} \Phi' \left( \frac{x}{\sqrt{T_0}} \right) d x. \end{align*}

From here perhaps you can substitute in your exact formulae for $d_1,\,d_2$ and simplify the above.

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  • $\begingroup$ One quick question, what do you mean by $W_{T_0} \in dx$? $\endgroup$
    – Geoff
    Commented Apr 11, 2018 at 13:26
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    $\begingroup$ Since $W_{T_0}$ is a continuous variable, $$\mathbf Q(W_{T_0} = x) = 0, \,x \in \mathbb R.$$ Instead, we work with probability density function, which give a measure of the variable being in an infinitesimally small neighborouhood of $x$.This is largely technical point, so for heuristic purposes you can think of $W_{T_0} \in dx$ as being the same as $W_{T_0} =x$ and condition on this. A discussion of this notation is given here, where it is picked up that although common, it is itself misleading. $\endgroup$
    – owen88
    Commented Apr 11, 2018 at 16:01
  • $\begingroup$ Are that makes sense, thank you! Also one last thing, I am a little confused as how you have used the law of total expectation? $\endgroup$
    – Geoff
    Commented Apr 12, 2018 at 8:51
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    $\begingroup$ Apologies, that should read law of total probability, this is now corrected. $\endgroup$
    – owen88
    Commented Apr 12, 2018 at 9:59
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    $\begingroup$ This continuous in the same vein as interpreting the condition $\{W_{T_0} \in dx \}$ as being $\{W_{T_0} = x\}$. If you make this heuristic substitution, then \begin{align} \{W_T > d_2 | W_{T_0} = x\} &= \{ W_T - W_{T_0} > d_2 - W_{T_0} \, | \, W_{T_0} = x\} \\ &= \{W_T - W_{T_0} > d_2 - x \, | \, W_{T_0} = x\} \\ &= \{W_T - W_{T_0} > d_2- x\},\end{align} where in the final equality we use independence of the interval $W_T - W_{T_0}$ from $W_{T_0}$. $\endgroup$
    – owen88
    Commented Apr 13, 2018 at 14:12

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