If $Y$ is lognormally distributed, what is $E|Y - 1|$? Let $Y$ be lognormal with parameters $\mu$ and $\sigma$, such that $\log(Y)$ is Gaussian with mean $\mu$ and variance $\sigma^2$.
I know $0 < Y < \infty$, $E(Y) = \exp(\mu+\frac{1}{2}\sigma^2)$, and $Var(Y) = \exp(2\mu +\sigma^{2})[\exp(\sigma^{2})-1]$
Let $R = Y - 1$.
$-1 < R < \infty$, $E(R) = \exp(\mu+\frac{1}{2}\sigma^2) - 1$, and $Var(R) = Var(Y) = \exp(2\mu +\sigma^{2})[\exp(\sigma^{2})-1]$
What is $E(|R|)$?
This is the mean of the "folded" distribution of $R$.
Context:
When you compare a measurement $M$ to a known control $C$, the quantity $\frac{M}{C}$ is often lognormally distributed.  $\frac{M-C}{C} = \frac{M}{C} - 1$ is the "relative difference" or RD.  In the above, $\frac{M}{C} = Y$ and $Y - 1 = R$ is the RD.  $E(R)$ is the "mean relative difference" or MRD.
$|R|$ is the "absolute relative difference" (not my choice of terms) or ARD, and $E(|R|)$ is the "mean absolute relative difference" or MARD.  This quantity is used to summarize the accuracy of a measurement system, especially the accuracy of blood glucose monitors.
 A: Continuing on from the previous answer.
We can derive the distribution function of your random variable, which will then allow us to derive its expectation.
Let's say we have
$$X\sim LN(\mu,\sigma^2)$$
Because we know the distribution of $X$, we know its distribution function, $F_{X}(x)$ where $0\leq x <\infty$.
Now we have that $Y=X-1$. We can define the cumulative distribution function of $Y$ as
$$F_{Y}(y)=\text{Pr}(Y\leq y)=\text{Pr}(X-1\leq y)=\text{Pr}(X\leq y+1)=F_{X}(y+1)$$
where $-1\leq y <\infty$.
Furthermore, we know that $Z=|Y|$. Applying a similar approach we get two cases.

For $-1\leq y<0\quad\Rightarrow\quad Z=-Y\quad\Rightarrow\quad 0\leq z<1$:

$$\begin{align}
F_{Z}(z)&=\text{Pr}(Z\leq z)\\
&=\text{Pr}(-Y\leq z)\\
&=1-\text{Pr}(Y\leq -z)\\
&=1-F_{Y}(-z)\\
&=1-F_{X}(-z+1)
\end{align}$$
$$\begin{align}
f_{Z}(z)&=\frac{d}{dz}F_{Z}(z)=f_{X}(-z+1)
\end{align}$$

For $0\leq y<\infty\quad\Rightarrow\quad Z=Y\quad\Rightarrow\quad 0\leq z<\infty$:

$$\begin{align}
F_{Z}(z)&=\text{Pr}(Z\leq z)\\
&=\text{Pr}(Y\leq z)\\
&=F_{Y}(z)\\
&=F_{X}(z+1)
\end{align}$$
$$\begin{align}
f_{Z}(z)&=\frac{d}{dz}F_{Z}(z)=f_{X}(z+1)
\end{align}$$
We then need to calculate the expectation of this random variable:
$$\begin{align}
\mathbb{E}[Z]&=\int_{0}^{1}z\cdot f_{X}(-z+1)dz+\int_{0}^{\infty}z\cdot f_{X}(z+1)dz\\
&=\int_{0}^{1}(1-j)f_{X}(j)dj+\int_{1}^{\infty}(i-1)f_{X}(i)di\quad\quad(j=-z+1, i=z+1)\\
&=\int_{0}^{1}f_{X}(j)dj-\int_{1}^{\infty}f_{X}(i)di-\int_{0}^{1}jf_{X}(j)dj+\int_{1}^{\infty}if_{X}(i)di\\
&=2\cdot F_{X}(1)-1-\int_{0}^{1}jf_{X}(j)dj+\int_{1}^{\infty}if_{X}(i)di\\
&=e^{\mu+\sigma^2/2}\Bigg(2\cdot\Phi\bigg(\frac{\mu}{\sigma}+\sigma\bigg)-1\Bigg)-\Bigg(2\cdot\Phi\bigg(\frac{\mu}{\sigma}\bigg)-1\Bigg)
\end{align}$$
where $\Phi(\cdot)$ is the cumulative distribution function of the standard Normal.

The two integrals above (in terms of $i,j$) can each be evaluated by using the two-sided truncated integral:
$$\begin{align}
\int_{a}^{b}\frac{1}{\sigma\sqrt{2\pi}}\exp\bigg(-\frac{(\ln{x}-\mu)^2}{2\sigma^2}\bigg)dx\\
\end{align}$$
Let $y=\ln{x}-\mu\quad\Rightarrow\quad dx=e^{y+\mu}dy$
$$\begin{align}
\Rightarrow e^{\mu+\sigma^2/2}\int_{\ln{a}-\mu}^{\ln{b}-\mu}\frac{1}{\sigma\sqrt{2\pi}}\exp\bigg(-\frac{(y-\sigma^2)^2}{2\sigma^2}\bigg)dy\\
\end{align}$$
Let $z=\frac{y-\sigma^2}{\sigma}\quad\Rightarrow\quad dy=\sigma dz$
$$\begin{align}
&\Rightarrow e^{\mu+\sigma^2/2}\int_{(\ln{a}-\mu-\sigma^2)/\sigma}^{(\ln{b}-\mu-\sigma^2)/\sigma}\frac{1}{\sqrt{2\pi}}\exp\bigg(-\frac{z^2}{2}\bigg)dz\\
&=e^{\mu+\sigma^2/2}\Bigg(\Phi\bigg(\frac{\ln{b}-\mu-\sigma^2}{\sigma}\bigg)-\Phi\bigg(\frac{\ln{a}-\mu-\sigma^2}{\sigma}\bigg)\Bigg)
\end{align}$$
A: According to Wolfram Mathematica, in general, for $a>0$, $\mathbb{E}_{x\sim\mathrm{N}(\mu,\sigma^2)}[|e^x - a|] $ is given by
$$
e^{\mu+\frac{\sigma^2}{2}}\,
\mathrm{Erf}\bigg(\frac{\mu+\sigma^2-\log a}{\sqrt{2\sigma^2}}\bigg) -
a\,
\mathrm{Erf}\bigg(\frac{\mu-\log a}{\sqrt{2\sigma^2}}\bigg).
$$
A: Thank you to epp and John Madden for your answers!
I am posting this as an answer rather than a comment because I have adapted and elaborated on epp's response, substituting my original variables and rearranging terms.
Let's say we have
$$Y\sim LN(\mu,\sigma^2)$$
$$F_{Y}(y) = \Phi \bigg(\frac{\log{y} - \mu}{\sigma}\bigg)  \quad 0\leq y < \infty$$
$$f_{Y}(y) = \phi \bigg(\frac{\log{y} - \mu}{\sigma}\bigg) \cdot \frac{1}{\sigma y}  \quad 0\leq y < \infty$$
where $\Phi(\cdot)$ and $\phi(\cdot)$ are the CDF and PDF of the standard Normal.
Now we have that $R=Y-1$.
\begin{align}
F_{R}(r)&=F_{Y}(r+1) \\
f_{R}(r) &= f_{Y}(r+1)   \quad -1 \leq r < \infty
\end{align}
Let $Z=|R|$.
\begin{align} 
f_{Z}(z) &= f_{Y}(-z+1) + f_{Y}(z+1) \quad 0 \leq z < 1 \\
f_{Z}(z) &= f_{Y}(z+1) \quad \quad z \geq 1
\end{align}
We then need to calculate the expectation of this random variable $Z$ (following epp's answer):
\begin{align}
\mathbb{E}[Z]&=\int_{0}^{1}z\cdot f_{Y}(-z+1)dz+\int_{0}^{\infty}z\cdot f_{Y}(z+1)dz\\
&=\int_{0}^{1}(1-j)f_{Y}(j)dj+\int_{1}^{\infty}(i-1)f_{Y}(i)di\quad\quad(j=-z+1, i=z+1)\\
&=\int_{0}^{1}f_{Y}(j)dj-\int_{1}^{\infty}f_{Y}(i)di-\int_{0}^{1}jf_{Y}(j)dj+\int_{1}^{\infty}if_{Y}(i)di\\
&=2\cdot F_{Y}(1)-1-\int_{0}^{1}jf_{Y}(j)dj+\int_{1}^{\infty}if_{Y}(i)di\\
\end{align}
epp's answer shows that
$$\int_{a}^{b}jf_{Y}(j)dj =   e^{\mu+\sigma^2/2}\Bigg(\Phi\bigg(\frac{\ln{b}-\mu-\sigma^2}{\sigma}\bigg)-\Phi\bigg(\frac{\ln{a}-\mu-\sigma^2}{\sigma}\bigg)\Bigg)
$$
So,
$$-\int_{0}^{1}jf_{Y}(j)dj+\int_{1}^{\infty}if_{Y}(i)di = e^{\mu+\sigma^2/2}\bigg(2\cdot\Phi\bigg(\frac{\mu}{\sigma} + \sigma \bigg) - 1\bigg)$$
Meanwhile,
$$2\cdot F_{Y}(1)-1= - \bigg( 2\cdot\Phi\bigg(\frac{\mu}{\sigma}\bigg) - 1\bigg)$$
Which leads to the final expression:
$$\mathbb{E}[Z] = e^{\mu+\sigma^2/2}\bigg(2\cdot\Phi\bigg(\frac{\mu}{\sigma} + \sigma \bigg) - 1\bigg) - \bigg( 2\cdot\Phi\bigg(\frac{\mu}{\sigma}\bigg) - 1\bigg)$$
This corresponds to John Madden's formula from Mathematica with $a = 1$:
$$
e^{\mu+\sigma^2/2}\,
\mathrm{Erf}\bigg(\frac{\mu+\sigma^2}{\sqrt{2\sigma^2}}\bigg) -
\mathrm{Erf}\bigg(\frac{\mu}{\sqrt{2\sigma^2}}\bigg).
$$
because
$$\mathrm{Erf}(x) = 2\cdot\Phi(x\sqrt{2}) - 1$$
