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If $X $ and $Y$ are independent random variables such that $X+Y$ has the same distribution as $X$ then is it always true that $P(Y=0)=1\ ?$

[This is actually a fact that a researcher used (without proof) while giving a lecture on his new paper that I was attending.]

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    $\begingroup$ This reads like a routine textbook exercise. Is it for some class? Consider the variance of the sum. What does it imply about $Y$? $\endgroup$ – Glen_b Nov 5 '16 at 23:49
  • $\begingroup$ @Glen_b In this problem ,is it right to assume that $E[X]$ and $Var[X]$ is finite? $\endgroup$ – Qwerty Nov 6 '16 at 5:15
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    $\begingroup$ This is actually a fact that a researcher used (without proof) while giving a lecture(I was attending) on his new paper. That's why I had not added the [self-study] tag. $\endgroup$ – Qwerty Nov 6 '16 at 5:20
  • $\begingroup$ In relation to the question about assuming finite moments (which is a good point) -- what's okay to assume about the problem is up to you -- it was a suggestive hint, not an answer. You say it's a research problem, so it's up to you to say what the conditions were, in the question. However, even if your problem is a more general one, the variance argument may give you a good sense of how to generalize it. [How does the research problem arise/what research area is this? Was it an application where the variables are bounded for example? Many real-world variables are.] $\endgroup$ – Glen_b Nov 6 '16 at 5:37
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    $\begingroup$ In the case of more general variables, one might perhaps move to using characteristic functions. ... in fact it's easy to do so. I have posted an answer. $\endgroup$ – Glen_b Nov 6 '16 at 5:56
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The variance argument isn't hard to make more general:

Consider the characteristic function of the sum $\phi_{X+Y}(t) = \phi_X(t) \phi_Y(t)$

But since $X$ and $X+Y$ have the same distribution $\phi_{X+Y}(t) = \phi_X(t)$.

Hence $\phi_Y(t) = \phi_{X+Y}(t) /\phi_X(t) = 1$

This is the characteristic function of a degenerate distribution with all its mass at $0$.

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    $\begingroup$ Oh, actually, isolated points shouldn't make a difference when we do the integration to take it back, so as long as $\phi_X$ is not zero on an interval/region it shouldn't matter. Anyway, I'll come back to it $\endgroup$ – Glen_b Nov 6 '16 at 6:27
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    $\begingroup$ @Glen_b: a simple argument completing yours is that $\phi_X(0)=\phi_Y(0)=1$ and that both functions are non-vanishing in a neighbourhood of zero due to the continuity of the characteristic function. $\endgroup$ – Xi'an Nov 6 '16 at 11:06
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    $\begingroup$ @Qwerty: I do not understand how missing the complete argument can prevent you from attending the next lecture. Of course, if this was an homework assignment and you had to return it by this deadline, it would explain for the urgency... $\endgroup$ – Xi'an Nov 6 '16 at 11:08
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    $\begingroup$ @Qwerty I'll have to try to do it properly later in an edit to the answer, bu I can outline the gist of it (you may have seen I hinted as such an argument in my first comment under my answer, before I edited it back out) $\phi_X$ must be non-vanishing in a neighborhood of $0$, so $\phi_Y$ must then be $1$ in that neighborhood of $0$. That would actually be enough to put all the probability for $Y$ at 0. (e.g. it means that all the moments must be zero.) $\endgroup$ – Glen_b Nov 6 '16 at 11:31
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    $\begingroup$ Oh, Xi'an has a better argument there. $\endgroup$ – Glen_b Nov 6 '16 at 11:32
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My approach. We are essentially comparing $X$ with $X+Y$.

The variance of $X+Y$ is, given independence:

$$\begin{align} \text{Var}(X+Y)&=\text{Var}(X)+\text{Var}(Y)\\ \end{align}$$

We know that $X$ has variance $\text{Var}(X)$, which implies that $\text{Var}(Y)=0$. What this tells us is that $Y$ is (almost surely?) a constant.

Now, taking the expectaton of $X+Y$:

$$\begin{align} \mathbb{E}[X+Y]&=\mathbb{E}[X]+\mathbb{E}[Y]\\ \end{align}$$

But we know that $X$ has expectation of $\mathbb{E}[X]$, which implies $\mathbb{E}[Y]=0$.

So, $Y$ is a constant with expectation zero. This implies that $\text{Pr}(Y=0)=1$.

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  • $\begingroup$ Nice answer, StatsPlease. But did you see the comments by Glen_b above? It was asked that they show their progress first. $\endgroup$ – Xu Wang Nov 6 '16 at 4:32
  • $\begingroup$ @XuWang Should I delete my answer? $\endgroup$ – epp Nov 6 '16 at 4:33
  • $\begingroup$ I don't know. I am new here too :) $\endgroup$ – Xu Wang Nov 6 '16 at 4:34
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    $\begingroup$ And I think, one thing you have assumed wrong. $E[X]+E[Y]=E[X]\implies E[Y]=0$ if and only $E[X]$ is finite. But nowher has it been said so!. Take an example the Cauchy distribution! $\endgroup$ – Qwerty Nov 6 '16 at 5:12
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    $\begingroup$ @StatsPlease You don't have to delete it now you've posted it. But be aware of our policy on routine bookwork questions for future reference. (See the self-study guidelines for example). $\endgroup$ – Glen_b Nov 6 '16 at 5:29

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