Timeline for Proof that joint probability density of independent random variables is equal to the product of marginal densities
Current License: CC BY-SA 3.0
7 events
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| Sep 10, 2017 at 13:47 | comment | added | Zen | @Tielfish Poele: in this argument, $f_{X\mid Y}(x\mid y)=f_X(x)$ follows from the first formula $f_{X,Y}(x,y)=f_{X\mid Y}(x\mid y)f_Y(y)$ if you use the fact that $f_{X,Y}(x,y)=f_X(x) f_Y(y)$ for independent $X$ and $Y$ (and take care of possible division by zero). But you can't use this fact, because you're trying to prove it. | |
| Sep 9, 2017 at 10:23 | comment | added | Stéphane Laurent | I disagree with "$f_{X,Y}(x,y) = f_{X\mid Y}(x\mid y)f_Y(y)$ by definition". The conditional density is a mathematical object of higher level than the joint density. | |
| Sep 9, 2017 at 3:08 | comment | added | Dale C | Really? He's trying to prove that a joint is the product of marginals under independence. What in that line is a joint distribution? | |
| Sep 9, 2017 at 2:31 | comment | added | Zen | I think the logic of this argument is not correct. When it says that $f_{X\mid Y}(x\mid y) = f_X(x)$ it's basically using what it's trying to prove. See the answer bellow. | |
| Sep 9, 2017 at 1:02 | vote | accept | jschnieder | ||
| Mar 15, 2018 at 23:04 | |||||
| Sep 9, 2017 at 0:37 | history | edited | Dale C | CC BY-SA 3.0 |
added 517 characters in body
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| Sep 9, 2017 at 0:27 | history | answered | Dale C | CC BY-SA 3.0 |