So I came across this exercise in my book.

a) Let $U$ be any random variable and $V$ be any other nonnegative random variable. Show that $$F_{U+V}(t) \le F_U(t)\ for\ every \ t $$

b)Describe formally the following model. Two groups of $n_1$ and $n_2$ individuals, respectively, are sampled at random from a very large population. Each member of the second (treatment) group is administered the same dose of a certain drug believed to lower blood pressure and the blood pressure is measured after 1 hour. Each member of the first (control) group is administered an equal dose of placebo and then has the blood pressure measured after 1 hour. It is known that the drug either has no effect or lowers blood pressure, but the distribution of blood pressure in the population sampled before and after administration of the drug is quite unknown.

So for the a) I showed that $F_U(t)-F_{U+V}(t)\ge0$ which led me to $\int_{t-v}^tf_U(t)dt \ge0$ as f is assumed to be a legitimate PDF hence $F_{U+V}(t) \le F_U(t)\ for\ every \ t $

$X_i$: Blood Pressure from the ith individual,

$Y_i$: Response to the medication on the ith individual

For b) I set a model $Y_i$ being the response from $X_i-\Delta$ with $i=0,1,2,...,n_1$

and $Y'_i$ being the response from placebo on $X_i$ with $i=0,1,2,...,n_2$

I could do the same thing as a) but I don't know if I should assume the $X's$ to be normal with mean $\mu$ and variance $\sigma^2$ which are unknowns.

Does anyone know how to go about that? Any input will be greatly appreciated.

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    $\begingroup$ You might find it instructive to prove part (a) without introducing the (limiting and unnecessary) assumption that $U$ has a density. As far as (b) goes, a similar observation holds: don't introduce assumptions that weren't made in the problem statement. $\endgroup$ – whuber Oct 24 at 21:41
  • $\begingroup$ I couldn't prove without doing so. Do you have tips? I know from googling about Stochastic ordering but couldn't use anything from it. $\endgroup$ – Mahamad A. Kanouté Oct 24 at 21:59
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    $\begingroup$ One proof of (a) observes that since $V\ge 0,$ the condition $U+V\le t$ implies (via the axioms of ordering) that $U\le t.$ An axiom of probability immediately tells us that $$F_{U+V}(t)=\Pr(U+V\le t)\le \Pr(U\le t)=F_U(t),$$ QED. $\endgroup$ – whuber Dec 8 at 23:08
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    $\begingroup$ I would be able to cross off this exercise now. I really appreciated this response, thank you ! @whuber $\endgroup$ – Mahamad A. Kanouté Dec 8 at 23:13

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