# showing 2 random variables having same probability mass function, then they have the same distribution

Given X and Y be the discrete random variables and having the same probability mass function. How do i show that these in fact have the same distribution?

I am just stuck on how to start this. I was thinking letting P be the probability mass function, F and G be the distribution for X and Y resp. Then we show that $F(x) = \sum_x{P(X = x)}$ and $G(y) = \sum_y{P(Y = y)}$. But then F and G may not be in the same space nor the set X and Y are defined in.

It might be that $X$ and $Y$ are defined on different probability spaces, but that is not relevant here.

If $\langle\Omega_1,\mathcal A_1,P_1\rangle$ and $\langle\Omega_2,\mathcal A_2,P_2\rangle$ are probability spaces, and $X:\Omega_1\to\mathbb R$, $Y:\Omega_2\to\mathbb R$ are Borel-measurable functions, then probability spaces $(\mathbb R,\mathcal B,P_X)$ and $(\mathbb R,\mathcal B,P_Y)$ are induced by $X$ and $Y$.

Here $\mathcal B$ denotes the collection of Borel-measurable sets, and $P_X$ is the so-called distribution of $X$ and is defined for $B\in\mathcal B$ by:$$P_X(B):=P_1(X\in B)=P_1(\{\omega\in\Omega_1\mid X(\omega)\in B\})$$

To be shown is that $P_X=P_Y$.

If $f_X$ denotes the PMF of $X$ then $P_X(B)=\sum_{x\in B\cap S}f_X(x)$ where $S$ is a countable support of $X$.

But we have $f_X=f_Y$ so:$$P_X(B)=\sum_{x\in B\cap S}f_X(x)=\sum_{x\in B\cap S}f_Y(x)=P_Y(B)$$

• thank you. this makes more sense Commented Jul 12, 2017 at 18:04
• You are very welcome. Commented Jul 12, 2017 at 18:05