Is the Monotone Likelihood Ratio Property (MLRP) preserved under mean-preserving spreads? I have a spent a long time trying to find an answer to this question, so any help would be much appreciated.
Let $X_1$ be a random variable (r.v.) with distribution $f_1$ and $X_2$ be a r.v. with distribution $f_2$. Suppose $f_1$ and $f_2$ satisfy the monotone likelihood ratio property (MLRP) such that
\begin{equation}
\frac{f_1(x_1)}{f_2(x_1)}\geq \frac{f_1(x_0)}{f_2(x_0)} ; \ \forall x_1\geq x_0 
\end{equation}
Let $Y_i = X_i + e_i $ be a mean preserving spread of $X_i$, with $e_i$ ~ $N(0,1)$, and with distribution $g_i(y)$, for $i=1,2$.  Does the MLRP hold for $g_1(y)$ and $g_2(y)$?  
 A: Here are some thoughts (not a clear cut answer, but it is too long for a comment).
For the general case when $X$ and $e$ are independent with densities that are designated by $f_X(x)$ and $f_E(e)$, thus for $Y = X + e$ you have that the density of $Y$ can be calculated using convolution 
$$
g(y) = \int_{-\infty}^{\infty}f_X(y-t)f_E(t)dt, 
$$
hence,
$$
\frac{g_1(y_1)}{g_2(y_1)} = \frac{\int_{-\infty}^{\infty}f_{X_1}(y_1-t)f_E(t)dt}{\int_{-\infty}^{\infty}f_{X_2}(y_1-t)f_E(t)dt} .
$$
I'm note sure that it is straightforward to claim that this inequality 
$$
\frac{f_{X_1}(x_1)}{f_{X_2}(x_1)} \ge \frac{f_{X_1}(x_0)}{f_{X_2}(x_0)}, 
$$ 
preserved under convolution, i.e., implies that
$$
\frac{\int_{-\infty}^{\infty}f_{X_1}(y_1-t)f_E(t)dt}{\int_{-\infty}^{\infty}f_{X_2}(y_1-t)f_E(t)dt} \ge  
\frac{\int_{-\infty}^{\infty}f_{X_1}(y_0-t)f_E(t)dt}{\int_{-\infty}^{\infty}f_{X_2}(y_0-t)f_E(t)dt}, \forall y_1 \ge y_0.
$$
We can inspect one very simple case. Let $X \sim N(\mu, \sigma^2)$ and $e \sim N(0,1)$, when they are independent r.vs, thus $X+e = Y \sim N(\mu, \sigma^2+1)$. In this case clearly the original monotonicity is preserved as we only enlarged the variance (dispersion) of each $X_i$. However, I'm not sure that every possible convolution will preserve such a nice property.   
