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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.

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The easier way is to use expectation forms for the whole formula, like following: \begin{equation} \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} \end{equation} Thus, you got the answer.

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