# SSD and Mean preserving spread

Consider the following two random variables $$X$$ and $$Y$$ for risky loss: $$Y$$ is exponentially distributed with parameter $$b = 0.005$$ (the density is $$f_Y(x) = be^{-by}$$) and $$X$$ is normally distributed with mean $$\mathbb{E}[Y]$$ and variance $$\sigma^2$$. We want to answer the following:

Under what conditions could we say that $$(w_0-Y)$$ second-order stochastically dominates $$(w_0-X)$$ and under what conditions could we say that $$(w_0-X)$$ second-order stochastically dominates $$(w_0-Y)$$? (The past exam question does not specify what is $$w_0$$, but I guess should be a constant representing initial wealth.)

My approach is as follows - we consider the equivalent condiition:

(And I hope here by $$f_1$$, $$f_2$$ they mean probability densities and not cdf-s; correct me if wrong, please)

So now we consider $$f_X(x) \geq f_Y(x)$$. Firstly, it holds for $$x<0$$, as exponential distribution has no support on $$x<0$$. Now, for $$x\geq 0$$ it is equivalent to

$$x^2 - 2(b^{-1} + \sigma^2b)x + b^{-2} + 2\sigma^2\log(b\sigma\sqrt{2\pi}) \leq 0$$

Now if $$b^{-2} + 2\sigma^2\log(b\sigma\sqrt{2\pi}) \leq 0$$, then the quadratic has two solutions $$x_1 \leq 0 < x_2$$ and the inequality overall holds for $$(-\infty, x_2]$$ (and the reverse one for $$[x_2, \infty]$$. So in this case neither dominates (or both dominate each other if this even makes sense? The thing is, both sides satisfy the $$I$$-interval condition)

And if $$b^{-2} + 2\sigma^2\log(b\sigma\sqrt{2\pi}) > 0$$ there are either two positive solutions $$x_1 < x_2$$ or no real solutions. In the former case the inequality overall holds for $$(-\infty,0) \cup [x_1, x_2]$$ and in the latter -- for $$(-\infty, 0)$$. In the first case the $$I$$-interval condition is satisfied from neither and in the second -- from both sides. So again neither dominates/both dominate?

Is this correct? Seems extremely strange to be asked on a (past) exam if it is like that... Thank you!