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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: enter image description here

(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!

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