The problem sites: we have two variables X and Y, where $X\sim Be(p)$ and $Y\sim Po(p)$. Create a probability space where:

$$\mathbb{P}[X\neq Y]\leq p^{2}$$

What I have tried so far:

Because X follows a Bernoulli distribution, it can only be either 0 or 1. However, Y follows a Poisson distribution, therefore it can be equal to any $n\in \mathbb{N}$.

I'll assume that we get pairs of (x,y). From the limitations I get:

$$ \mathbb{P}[X\neq Y] = 1-\mathbb{P}[X=Y] = 1-\mathbb{P}[(0,0)]-\mathbb{P}[(1,1)] \leq p^{2}$$

From that I get that $p\geq 0.78$.

I'm not sure I've thought this correctly or not. If I'm correct, where should I go from where I'm left off. If not, what am I doing wrong?


1 Answer 1


I might be wrong, but I think the problem is asking for something different.

Basically, to phrase it differently, what they are asking you is to "create" or define a joint distribution of $(X,Y)$ such that for all (sensible) $p$ that inequality holds.

To maybe make this more concrete on why I think that's what they are asking:

How did you calculate $\mathbb{P}((X,Y) = (0,0))$ and $\mathbb{P}((X,Y) = (1,1))$? You have no information given about their joint distribution and that's, as far as I can tell, exactly the point of this question. If you just calculated it as if they are independent, you are making an assumption on exactly that: their joint distribution. But there are various other ways, how $X$ and $Y$ might "relate". And you have to find one "suitable" way (i.e. such that the inequality holds).

  • $\begingroup$ Yes, in my assumption I said that they're independent. For that, I said if $\mathbb{P}[X=0]=1-p$ I get $\mathbb{P}[(0,0)]=\mathbb{P}[X=0]\mathbb{P}[Y=0]=(1-p)\frac{p^{0}e^{-p}}{0!}$ and similarly for if $X=Y=1$. I got the result for p as stated in my question after making sure the limited phrase was satisfied, and also checked if for every pairing $(0,y) \mbox{ or } (1,y)$ the probability is $p_i\in[0,1]$ and if $\mathbb{P}[\Omega]$ are true, which they were. So I said that is my probability space. Is that correct? Are there more I could create? $\endgroup$
    – Tita
    Commented Nov 15, 2021 at 19:35
  • 1
    $\begingroup$ As phrased, I think one should suffice. However, it doesn't actually work for all $p \in (0,1]$ if you model them to be independent: If $X,Y$ are independent, then I get $\mathbb{P}(X \neq Y) = 1 - \frac{3}{4} \cdot e^{-1/2} > \frac{1}{4} = p^2$ for $p = \frac{1}{2}$. $\endgroup$
    – MXXZ
    Commented Nov 16, 2021 at 12:12

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