Expected value of max(R-c,0)

Question

Let $R$ bet Log-Normally distributed i.e. LN($\mu,\sigma^2$).

Now we want to find the expected value:

$\mathbf{E}[\max(R-c,0)]$

and also second order moment:

$\mathbf{E}[\max(R-c,0)^2]$.

I have trouble in understanding exactly how should I intrepete the max function in the expectation formula, because it should be something like:

$\mathbf{E}[\max(R-c,0)]=\int_{-\infty}^{+\infty}\max(R-c,0)f(z)dz=\int_{-\infty}^{+\infty}\max(e^{\mu+\sigma z}-c,0)f(z)$dz where z is N(0,1).

In the solutions they are using indicator functions but I don't understand why they use them. Anyone can help me out? Thanks

Answer 1

If $c\le0$ then $\max\{R-c,0\}=R-c$. So, I am taking $c>0$. Then, the first integration becomes $$I=\int_{a}^{\infty}e^{\mu+\sigma z}f(z) dz$$ where $a=\dfrac{\ln c-\mu}{\sigma}$