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When we write $X\stackrel d= Y$, does this mean that $X$ and $Y$ have exactly the same distribution?

For example, $X\sim\mathcal N(\mu,\sigma^{2}) \ \text{and}\ X\stackrel d= Y\implies Y\sim\mathcal N(\mu,\sigma^{2})$ with the same values of $\mu$ and $\sigma$?

From what I understand in this Wikipedia article, $X\stackrel d= Y\implies F_X(t)=F_Y(t)$ for all $t$, which would seem to imply the answer to my question is yes. But then it goes on to say that equality of distribution is the weakest form of equality/convergence usually discussed. What about this type of equality makes it weak?

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$X\overset{d}{=}Y$ does mean equality in distribution, which implies that the distribution functions of $X$ and $Y$ agree, that is, $$ F_X(t)=F_Y(t),\quad\forall t\in\Bbb R. $$ Note that equality in distribution also sometimes goes by the name equality in law.

Since the notation $X\sim\mathcal N(\mu,\sigma^2)$ is just a statement about the distribution of $X$, you can rightfully conclude $$ X\overset{d}{=}Y\,\land\, X\sim\mathcal N(\mu,\sigma^2)\implies Y\sim\mathcal N(\mu,\sigma^2). $$

In regards to the strength (or weakness) of such an equality, note that there are multiple ways we can describe the "equalness" of two random variables. The reason why equality in distribution is considered weak is because it does not imply actual observed values of $X$ and $Y$ will be similar or close in value.

For example, suppose $X\sim\operatorname{Uniform}(-1/2,1/2)$ and $Y=-X$. Then by change of variables (or just upon inspection) we can see that $Y\sim\operatorname{Uniform}(-1/2,1/2)$ and thus $X\overset{d}{=}Y$. But if you were to then draw observations from these distributions they would never by similar as they would always be opposite in sign.

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