# Loss Size Index Function of A Lognormal Random Variable

I have this tutorial question and I've gone through the solutions, getting all but one line of working. I broke down the question to this point but I can't seem to get out the following.

So Loss Size index function at a point $d$ for variable $Y$ (range $0$ to infinity) in my course is defined as:

$$\frac{\int_{0}^{d}yf(y)\,dy}{E(Y)}$$

Now if $Y$ is lognormal, i.e. $\ln Y \sim N(\mu, \sigma^2)$ I can see that the loss size index function equals the following (but I want to know how to show it).

$\text{Pr}[ N < ( \ln d - (\mu + \sigma^2) ) / \sigma ]$ where $N$ is the standard normal.

While solving the "big" question I had to find $\text{Pr}[Y < d]$ which I managed to get out to be $\text{Pr}[ N < ( \ln d - \mu ) / \sigma ]$ after transforming the variable.

So I think I am struggling with working with the whole "expectation of lognormal" to get something that looks like the standard normal distribution function.

The easier way is to use expectation forms for the whole formula, like following: $$$$\begin{split} I & = \frac{\int_0^dy\mathrm{f}(y)\mathrm{d}(y)}{E(Y)}\\ & = \int_0^d \frac{1}{\sqrt{2\pi}\sigma} \frac{1}{y} exp\left\{\mathrm{log}y\right\} exp \left \{-\frac{(\mathrm{log}y - \mu)^2}{2\sigma^2}\right\}\mathrm{d}y \cdot exp\left\{-\mu-\frac{\sigma^2}{2}\right\}\\ & = \int_0^d \frac{1}{\sqrt{2\pi}\sigma} \frac{1}{y} exp \left \{-\frac{(\mathrm{log}y - \mu)^2-2\sigma^2\mathrm{log}y+2\sigma^2\mu+\sigma^4}{2\sigma^2}\right\}\mathrm{d}y\\ & = \int_0^d \frac{1}{\sqrt{2\pi}\sigma} \frac{1}{y} exp \left \{-\frac{(\mathrm{log}y)^2-2(\mu+\sigma^2)\mathrm{log}y+(\mu+\sigma^2)^2}{2\sigma^2}\right\}\mathrm{d}y\\ & = \int_0^d \frac{1}{\sqrt{2\pi}\sigma} \frac{1}{y} exp \left \{-\frac{(\mathrm{log}y-(\mu+\sigma^2))^2}{2\sigma^2}\right\}\mathrm{d}y\\ & = \Phi\left(\frac{\mathrm{log}y-(\mu+\sigma^2)}{\sigma}\right) \end{split}$$$$ Thus, you got the answer.