2
$\begingroup$

Let's say we have two random variables $X$ and $Y$ both defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$. So $X$ and $Y$ are measurable functions from $\Omega$ to $\mathbb{R}$.

Do $X$ and $Y$ necessarily have the same distribution? Can we for example have $X\sim Bern(p)$ and $Y\sim N(0,1)$? Or would we then say that $X$ and $Y$ have for example the measure $\mathbb{P}_X$ and $\mathbb{P}_Y$ respectively?

$\endgroup$

3 Answers 3

2
$\begingroup$

You can easily construct such $X$ and $Y$. Take $(\Omega,\mathcal{F},\mathsf{P})=((0,1),\mathcal{B}_{(0,1)},\lambda)$, where $\lambda$ is the Lebesgue measure, and set $$ X(\omega):=1\{\omega\le p\}, $$ and $$ Y(\omega):=\Phi^{-1}(\omega), $$ where $\Phi$ is the standard normal cdf. Then $$ \mathsf{P}(X=1)=\lambda(\{\omega\le p\})=p,\quad \mathsf{P}(X=0)=1-p, $$ and $$ \mathsf{P}(Y\le y)=\lambda(\{\Phi^{-1}(\omega)\le y\})=\lambda(\{\omega\le \Phi(y)\})=\Phi(y). $$ That is, $X\sim\text{Bern}(p)$ and $Y\sim N(0,1)$.


An interesting related question is whether one can construct independent random variables on the same probability space (s.t. $P_{X,Y}=P_X\otimes P_Y$).

$\endgroup$
1
$\begingroup$

No, they do not. For example, the random variable on the interval with the uniform measure which maps everything to $1$ has a delta function distiribution, while the one that maps $x$ to $x$ has uniform distribution, while the one that maps $x$ to $x^2$ has some non-uniform distribution.

$\endgroup$
1
$\begingroup$

Simple example: Two coins, one biased $P(heads)=.75$ and the other not $P(heads)=.5$. Same sample spaces for both.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .