If $X$ is log-normal, find $\mathbb{E}(\min(X-4,1))$. For a random variable $X$, we have $\mathbb{E}(X)=4$ and that $\ln X$ follows normal distribution with variance $3$.
I need to find $\mathbb{E}(\min(X-4,1))$ and express it with $\Phi$, the cumulative distribution function of the standard normal distribution. My attempt:
$$
\begin{align}
\mathbb{E}(\min(X-4,1))
&=\int^\infty_{-\infty} \min(x-4,1)f_X(x)dx\\
&=\int^5_{-\infty}(x-4)f_X(x)dx+\int^\infty_5f_X(x)dx\\
\end{align}
$$
Here's where I'm stuck and i don't know what to do to express it with $\Phi$, the CDF of the standard normal distribution. Any suggestions please?
 A: So we know $Y = \ln X \sim N(\mu,\sigma^2)$, so $X$ is a log-normal random variable. The mean of a log-normal random variable whose natural log has mean $\mu$ and standard deviation $\sigma$ is given by:
$$E[X]=\exp\left(\mu + \frac{\sigma^2}{2}\right)$$
We are told $E[X]=4$, and $Var[\ln X] =3 = \sigma^2$ therefore
$$E[X]=4 \implies \ln 4 = \mu + \frac32 \implies \mu = \ln 4 - \frac32$$
Using the standard normal CDF, we can write the cdf of $Y$ ($F_Y$) in terms of $\Phi$:
$$F_Y(y) = \Phi\left(\frac{y-\ln 4 + \frac32}{\sqrt{3}}\right)$$
Writing out the expectation
$$E[\min(X-4,1)] = \int_5^{\infty} f_X(z)dz + \int_0^5(z-4)f_X(z)dz = $$
$$\int_0^5zf_X(z)dz - \int_0^5 4f_X(z)dz + \int_5^{\infty} f_X(z)dz$$
The last term is just $P(X \geq 5)$, which, given that $\ln$ is a monotonic transformation of $Y \to \ln(Y)$ implies:
$$P(X\geq 5) = P(Y \geq \ln 5) = 1-\Phi\left(\frac{\ln 5-\ln 4 + \frac32}{\sqrt{3}}\right)$$
Similarly, the middle term is just $4P(X\leq 5)$, which is
$$4P(X\leq 5) = 4\Phi\left(\frac{\ln 5-\ln 4 + \frac32}{\sqrt{3}}\right)$$
Finally, we get to the first term:
$$E[X] = \int_0^5 zf_X(z)dz$$
This one is a little tricky -- there may be a slick way to get to this but I went with using the formula for the partial expectation of a lognormal random variable(and here, page 4)
$$E\left[X\mathbb{1}_{\leq k}\right] = \int_0^{k}zf_X(z)dz = \exp\left(\mu + \frac{\sigma^2}{2}\right)\Phi\left(\frac{\ln k -\mu - \sigma^2}{\sigma}\right)=E[X]\Phi\left(\frac{\ln k -\mu - \sigma^2}{\sigma}\right)=4\Phi\left(\frac{\ln k -\mu - \sigma^2}{\sigma}\right)$$
Putting it all together:
$$E[\min(X-4,1)] = 4\Phi\left(\frac{\ln 5 -\ln 4 + \frac32 - 3}{\sqrt{3}}\right)-4\Phi\left(\frac{\ln 5-\ln 4 + \frac32}{\sqrt{3}}\right)+1-\Phi\left(\frac{\ln 5-\ln 4 + \frac32}{\sqrt{3}}\right) = $$
$$1+4\Phi\left(\frac{\ln 5 -\ln 4 + \frac32 - 3}{\sqrt{3}}\right) -5\Phi\left(\frac{\ln 5-\ln 4 + \frac32}{\sqrt{3}}\right)\approx -2.28$$

R-code to validate
mu <- log(4)-3/2
sd <- sqrt(3)
x <- rlnorm(n=50000,meanlog=mu, sdlog=sd)
y <- pmin(x-4,rep(1,length(x)))
mean(y)
[1] -2.277581

