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Does the marginal pdf for X, $$\ f_X(x) = \int_{-\infty}^{\infty} f(x,y) \, dy$$ give us a probability if I put x in it? If so, I couldn't really understand. Because I thought that $\ f(x, y)$ does not give$\ P(X=x, Y=y)$. But here, I thought that $\ f_X(x)$ means $\ P(X=x, -\infty\le Y\le\infty)$ , which is not a region. Also, if I calculate $\ P(X=x, -\infty\le Y\le\infty)$ in a way of pdf defined, then $$\int_{x}^{x} \int_{-\infty}^{\infty} f(u,v) \,dv\,du$$ which will be $0$. I don't really understand this.

whwjddnjs
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    If this is a continuous distribution, then $P(X=x, Y=y)$ and $P(X=x, -\infty\le Y\le\infty)$ are both zero. Meanwhile $f(x,y)$ is the joint density and $f_X(x)$ the marginal density for $X$ – Henry Nov 21 '18 at 14:29
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    PDF's (marginal or not) do not "give" probabilities. As you said $f(x,y)$ cannot be identified as $P(X=x,Y=y)$. Similarly $f_X(x)$ cannot be identified as $P(X=x)$. – drhab Nov 21 '18 at 14:32

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As the name "probability density function" (abbreviated as "PDF") indicates, the values of such a function are not probabilities but probability densities. That is, if $g$ is the PDF of a random variable $X$, then the probability that $X$ is in a small interval around the point $x$ is approximately the product of $g(x)$ times the length of the interval. In other words, $g(x)$ is the probability per unit length ("probability density") near $x$.

In more precise mathematical language, $g(x)$ is the unique number (if one exists at all) such that, for every $\epsilon>0$, there exists $\delta>0$ such that, for every interval $I$ that contains $x$ and has length $<\delta$, $$ \left|\frac{P(X\in I)}{\text{length}(I)}-g(x)\right|<\epsilon. $$

In less precise (but perhaps more understandable) language, the probabiity that $X$ lies in an interval around $x$ of infinitesimal length $dx$ is $g(x)\,dx$.

Andreas Blass
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