Joint pdf as a product of two independent functions with dependent domain

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Let us have two random variables $X$ and $Y$ and their joint probability density function $f_XY(x,y)$. Let us assume that the function $f_XY$ can be written as a product of two functions $g(x)$ and $h(y)$. Then we say that the random variables $X$ and $Y$ are independent. But what if the range of $y$ in the domain of $f_XY$ is dependent of $x$? For example $f_XY$ is defined on the set $y$, are $X$ and $Y$ still independent?




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    Let us have two random variables $X$ and $Y$ and their joint probability density function $f_XY(x,y)$. Let us assume that the function $f_XY$ can be written as a product of two functions $g(x)$ and $h(y)$. Then we say that the random variables $X$ and $Y$ are independent. But what if the range of $y$ in the domain of $f_XY$ is dependent of $x$? For example $f_XY$ is defined on the set $y$, are $X$ and $Y$ still independent?




    probability probability-theory




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      Let us have two random variables $X$ and $Y$ and their joint probability density function $f_XY(x,y)$. Let us assume that the function $f_XY$ can be written as a product of two functions $g(x)$ and $h(y)$. Then we say that the random variables $X$ and $Y$ are independent. But what if the range of $y$ in the domain of $f_XY$ is dependent of $x$? For example $f_XY$ is defined on the set $y$, are $X$ and $Y$ still independent?




      probability probability-theory




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      Let us have two random variables $X$ and $Y$ and their joint probability density function $f_XY(x,y)$. Let us assume that the function $f_XY$ can be written as a product of two functions $g(x)$ and $h(y)$. Then we say that the random variables $X$ and $Y$ are independent. But what if the range of $y$ in the domain of $f_XY$ is dependent of $x$? For example $f_XY$ is defined on the set $y$, are $X$ and $Y$ still independent?





      probability probability-theory





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      asked 2 days ago









      Vwann

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          You would write your joint density $f_XY$ as
          beginalign*
          f_XY(x,y) = g(x)h(y) textbf1_leq 1, ,
          endalign*
          which can no longer be split up into a product of functions depending only on one variable (since you can't factor the indicator into two indicators).






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          • I see, thank you!
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          You would write your joint density $f_XY$ as
          beginalign*
          f_XY(x,y) = g(x)h(y) textbf1_leq 1, ,
          endalign*
          which can no longer be split up into a product of functions depending only on one variable (since you can't factor the indicator into two indicators).






          share|cite|improve this answer





















          • I see, thank you!
            – Vwann
            2 days ago














          up vote
          1
          down vote



          accepted










          You would write your joint density $f_XY$ as
          beginalign*
          f_XY(x,y) = g(x)h(y) textbf1_leq 1, ,
          endalign*
          which can no longer be split up into a product of functions depending only on one variable (since you can't factor the indicator into two indicators).






          share|cite|improve this answer





















          • I see, thank you!
            – Vwann
            2 days ago












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          You would write your joint density $f_XY$ as
          beginalign*
          f_XY(x,y) = g(x)h(y) textbf1_leq 1, ,
          endalign*
          which can no longer be split up into a product of functions depending only on one variable (since you can't factor the indicator into two indicators).






          share|cite|improve this answer













          You would write your joint density $f_XY$ as
          beginalign*
          f_XY(x,y) = g(x)h(y) textbf1_leq 1, ,
          endalign*
          which can no longer be split up into a product of functions depending only on one variable (since you can't factor the indicator into two indicators).







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          share|cite|improve this answer











          answered 2 days ago









          Daniel Xiang

          1,788413




          1,788413











          • I see, thank you!
            – Vwann
            2 days ago
















          • I see, thank you!
            – Vwann
            2 days ago















          I see, thank you!
          – Vwann
          2 days ago




          I see, thank you!
          – Vwann
          2 days ago












           

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