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Good evening, I need help with the resolution of the following exercise. I have two r.v. $X,Y$ that have laws $Px$ and $Py$ respectively. I Define $A(x)=P_X((-∞,x])+P_Y((-∞,x])$, now I have to verify some point.

  1. $A(x)$ is a decreasing function. I think is not since the higher the interval the higher the value of $A(x)$.
  2. $A(x)$ is always invertile. I think is not because it isn't a bijective function but I don't know if I'm right.
  3. there exits a probability $P$ on the event space $(R,B)$ where $B$ is the Borel sigma-algebra, s.t. $P((c,d])=A(d)-A(c)$ but I don't know how to show if it is correct or not
  4. $inf(x ∈ R: A(x)>=0.01)$ never exists, again I don't know how to procede.

I hope someone will help me, thank you anyway.

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    1) Yes you can show that $A$ is non-decreasing as it is the sum of two CDFs. 2) If the union of the support of $X, Y$ is not the whole real line, then are some time where the CDF of both are flat (constant). That's why we use the term non-decreasing instead of strictly increasing for the general case. 3) In general that should not be true. The value of $A$ can exceed 1. 4) In general wrong. If $A$ is continuous then it is done. If not, we know that it will jump over $0.01$ at some point which should be finite (unless $X, Y$ allow to have discrete mass at infinity) – 2017-02-04
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    First of all, thank you. Now the first point is very clear. The third one is also clear. I don't understand the second one, how can I show that A(x) is or is not invertible? The last one it's the most problematic to me... – 2017-02-04
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    To show that some statement is not always true, we construct a counter example against that. So you just need to construct a (simple) $A$ which is not always strictly increasing, i.e. constant at some interval in between, then you can argue that $A$ is not one-to-one/invertible. – 2017-02-06
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    Hi, thank you for the hint, I show you my idea: If A(x) is given by the sum of a discrete law and a continuos one, i.e. Is a mixture, therefore it will not be strictly increasing and not invertible. – 2017-02-06
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    Can you give me some advice for point 4? I really don't find a way to "escape" it. – 2017-02-06
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    again you just construct a simple example in which the infimum exist, say a continuous $A$. And as I said it maybe even more difficult to construct an $A$ which the infimum does not exist... – 2017-02-06
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    For example, if I construct A(x) as the sum of two absolutely continuous CDFs I'm done, right? Thanks for your help. – 2017-02-06

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