Second order stochastic dominance + mean preserving spread Let $X$ be a random variable and $F(X)$ associated distribution, and Let $Y = \alpha X +(1-\alpha),\alpha \in (0,1)$ and $G(Y)$ associated distribution. By linearity of expectation, $E[Y] = E[X]$, and by property of variance $Var(Y) = \alpha^2 Var(X)<Var(X) $ as $\alpha \in (0,1)$.
To me, it looks like $X$ is mean preserving spread of $Y$, and my question is that "Does $G(Y)$ second order stochastic dominates $F(X)$?
 A: So mean preserving spread $\implies$ SOSD. However, $E(X)=E(Y)$ AND $Var(X) > Var(Y) \nRightarrow Y$ SOSD $X$ (see here). However, an equivalent definition of SOSD is the following:
$Y$ SOSD $X$ iff for all $X$: $$\int_a^X F(x)dx \geq \int_a^X G(y)dy$$
where $F(a) = G(a) = 0$
Now consider your case (a little modified): $Y=aX+(1-a)\mu$, where $\mu=E(X)$ and $a\in (0,1)$. So we have that
\begin{align}
G(y) &= \mathbb P(Y \leq y) \\
&=\mathbb P(aX+(1-a)\mu \leq y) \\
&=\mathbb P\bigg(X \leq \frac{y-(1-a)\mu}{a}\bigg) \\
&=F\bigg(\frac{y-(1-a)\mu}{a}\bigg)
\end{align}
Now let $X$ (and so consequently $Y$) have finite support $[b,c]$ and define $S_f(x) = \int_b^xF(z)dz$ and $S_g(x) = \int_b^yG(z)dz$.
\begin{align}
S_g(y) &= \int_b^y G(z)dz \\
&=\int _b^y F\bigg(\frac{z-(1-a)\mu}{a}\bigg) dz \\
&=a\int _{b_1}^{y_1} F(\theta) d\theta \tag{$\theta =\frac{z-(1-a)\mu}{a}$ }
\end{align}
Now see that, $$\mu \geq b \implies  b \geq b_1\equiv \frac{b-(1-a)\mu}{a} , \forall a\in(0,1)$$
Therefore we have that:
\begin{align}
S_g(y) &=a\int _{b}^{y_1} F(\theta) d\theta \\
&=aS_f(y_1)
\end{align}
where $y_1 = (y-(1-a)\mu)/a$
Now consider the region where $c>y>\mu$. It's easy to see that $y_1>y$. Further, since $F(.)$ is a positive increasing function, $S_f(.)$ is also increasing. Also, let $1>a>S_f(y)/S_f(y_1)$
Therefore, for $y>\mu$,$$S_g(y) = aS_f(y_1) > S_f(y)$$
Now,  This proves, that $Y$ does not SOSD $X$, at least not for all values of $a$.
