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Let X be a real random variable with the density $f_X$ with respect to lebesgue measure given as…,which means $$P_X=f_X \cdot m$$ is a uniform distribution on the interval (0,1). What does it mean when $P_X$ is written that way? Is it a pushforward measure of a lebesgue measure or something else?(m is lebesgue measure)

Thanks in advance

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  • $\begingroup$ $X \colon \Omega \to \mathbb{R}$ and there is a probability $P$ on $\Omega$. Then $P_X$ is the pushforward measure of $P$ under the function $X$. This is the law (the distribution) of $X$. The notation $f_X\cdot m$ denotes the measure absolutely continuous with respect to $m$ whose Radon-Nikodym derivative is $f_X$. That is to say, $(f_X \cdot m)(A) = \int_Af_X \mathrm{d}m$. $\endgroup$ Commented Feb 7, 2024 at 17:38

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$P_X(A)=P(X^{-1}(A))$ for any Borel set $A$. $P_X$ is a probability measure on Borel sets of $\mathbb R$. The given equation says $P_X(A)=\int_A f_Xdm$ where $m$ is Lebesgue measure on $\mathbb R$.

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