# Solve a statistical equation

I got the following exercise where i have to understand the steps theres are done:

Since $z_T \sim N(0,T)$:

$$\text{Prob}((\mu-0.5\sigma^2)T+\sigma z_T>rT)$$ $$\text{Prob}(z_T>-\dfrac{(\mu-r-0.5\sigma^2)T}{\sigma})$$ $$\text{Prob}(z_T<\dfrac{(\mu-r-0.5\sigma^2)T}{\sigma})$$

Then i don't understand how i get to this:

$$\text{N}(\dfrac{(\mu-r-0.5\sigma^2)T}{\sigma\cdot\sqrt{T}})$$ $$\text{N}(\dfrac{(\mu-r-0.5\sigma^2)\sqrt{T}}{\sigma})$$

I hope someone can help - and thanks for your time!

• You seem to have dropped an $r$ from step 1 to step 2...also is $T$ supposed to be inside of the probability statement?? Commented Mar 30, 2014 at 14:42
• Sorry that i've forgot to add the r paranteses. It's done now :-) Commented Mar 30, 2014 at 14:45

I think the notation is poor but it looks like the last two lines are evaluating the inner term at the CDF of a normal. So from above we have that $z_T/\sqrt{T} \sim N(0,1)$, a standard normal random variable. So therefore
\begin{align*} \text{Pr}\left\{z_t < \dfrac{(\mu-r-0.5\sigma^2)T}{\sigma}\right\} &= \text{Pr}\left\{\dfrac{z_T}{\sqrt{T}} < \dfrac{(\mu-r-0.5\sigma^2)T}{\sigma\sqrt{T}}\right\} \\ &= \text{Pr}\left\{Z < \dfrac{(\mu-r-0.5\sigma^2)T}{\sigma\sqrt{T}}\right\} \\ &= \Phi\left(\dfrac{(\mu-r-0.5\sigma^2)T}{\sigma\sqrt{T}}\right) \end{align*}
Where $Z\sim N(0,1)$, a standard normal, and $\Phi$ is the standard normal CDF (i.e. $\Phi(z) = \text{Pr}\{Z < z\}$). This is more standard notation. In R you can find this probability with:
pnorm((mu-r-0.5*sigma^2)*T/(sigma*sqrt(T)), 0, 1)