Suppose $X_1$, $Y_1$, $X_2$ and $Y_2$ random variables with $\mathbb{N}$ (including $0$) as common support such that:

  • $X_1$ stochastically dominates $X_2$: $\Pr(X_1\leq x)\leq\Pr(X_2\leq x)$ for all $x\geq 0$. In fact, we're given a stronger condition: $\frac{\Pr(X_1=x)}{\Pr(X_2=x)}$ increases in $x$ (which implies stochastic dominance) but I'm not sure if this stronger fact is needed.
  • Conditioning on $X_i=x$, the CDF of $Y_i$ is $F_x$ (i.e. $Y_1|X_1=x$ and $Y_2|X_2=x$ share the same CDF $F_x$ which is parametrized by $x$).

Is the following true: $$ S_1\equiv X_1+Y_1\text{ stochastically dominates }S_2\equiv X_2+Y_2.\tag{$*$} $$ My intuition says YES but I cannot prove it.

In fact, I only got as far as: $$ \Pr[S_i\leq s]=\sum_{x=0}^sF_x(s-x)\Pr[X_i=x] $$ and so, for $s\geq 0$, $$ \Pr[S_1\leq s]-\Pr[S_2\leq s]=\sum_{x=0}^sF_x(s-x)\{\Pr[X_1=x]-\Pr[X_2=x]\}. $$


Without the monotonicity of the conditional probability, this is (perhaps surprisingly) not true.

Let $X_1 = 1$, $X_2 = 0$, and let $Y_1 = Y_2 = 5$ if $X_i = 0$, and $0$ otherwise.

With the monotonicity of the conditional probability, well, unless I am misinterpreting the condition, the property you are describing seems to imply that $X$ and $Y$ are independent. To see it, let $p_{i,j} = P(Y = i \mid X = j)$. Then, you are telling me that $p_{i,j+1} \geq p_{i,j}$ for all $i,j$? The problem occurs when you sum this up over all $i$, which gives you $\sum_i p_{i,j+1} \geq \sum_i p_{i,j}$. Now, both the left hand side and the right hand side are definitely 1, in which case none of the inequalities in $p_{i,j+1} \geq p_{i,j}$ could have been strict, i.e. they are all equalities, i.e. $p_{i,j}$ does not depend on $j$, i.e $P(Y = i) = P(Y = i \mid X = j)$, i.e. $Y$ is independent of $X$.

If you have that $X$ and $Y$ are independent, then yes, the stochastic dominance will be preserved. (Easier to see by just drawing out the convolutions).

  • $\begingroup$ Hi, you can’t sum $p_{ij}$ because for each $x$ that is conditioned on, $Y_i$ has a different distribution. $\endgroup$ – yurnero May 7 '18 at 13:18
  • $\begingroup$ Why can't I sum over the common support? sure, $p_{i,j}$'s can be different for each $i$ and $j$ but that should not be stopping me from summing them up? $\endgroup$ – E-A May 9 '18 at 20:42
  • $\begingroup$ You're right. There's something wrong with my formulation of the problem then. I apologize but would you mind deleting your answer so that I can delete the question and attempt to reformulate it. $\endgroup$ – yurnero May 10 '18 at 18:32
  • $\begingroup$ Uh, can you just accept this as the answer and ask a new question (perhaps referencing this one)? One might be able to use the answer given here as a reference while answering the new question. $\endgroup$ – E-A May 10 '18 at 21:20
  • $\begingroup$ Sorry E-A. I don't want to have a question under my record that is sub-par. Thanks to you, I've learned that I've erred somehow in my formulation and as a result, the question is a bit silly: in my context, $X_i$ and $Y_i$ definitely are not independent. I don't have time to fix the formulation at the moment and I'd rather not have the question published anymore until I have time to fix it. I hope you understand and help me. $\endgroup$ – yurnero May 10 '18 at 21:28

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