Why are the rows of the table linearly dependent for two independent RVs?
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I don't understand why the rows of this joint-distribution table are dependent for two, independent RVs.
What's the intuition here?
probability-distributions random-variables
asked Dec 20 '18 at 3:12
mitmath514mitmath514
213
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add a comment |
$begingroup$
I don't understand why the rows of this joint-distribution table are dependent for two, independent RVs.
What's the intuition here?
probability-distributions random-variables
asked Dec 20 '18 at 3:12
mitmath514mitmath514
213
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2
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Hint: $X$ and $Y$ are independent iff the joint density is the product of the marginal densities.
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– Math1000
Dec 20 '18 at 3:33
add a comment |
$begingroup$
I don't understand why the rows of this joint-distribution table are dependent for two, independent RVs.
What's the intuition here?
probability-distributions random-variables
asked Dec 20 '18 at 3:12
mitmath514mitmath514
213
$endgroup$
I don't understand why the rows of this joint-distribution table are dependent for two, independent RVs.
What's the intuition here?
probability-distributions random-variables
probability-distributions random-variables
asked Dec 20 '18 at 3:12
mitmath514mitmath514
213
asked Dec 20 '18 at 3:12
mitmath514mitmath514
213
asked Dec 20 '18 at 3:12
mitmath514mitmath514
213
asked Dec 20 '18 at 3:12
mitmath514mitmath514
213
asked Dec 20 '18 at 3:12
mitmath514mitmath514
213
213
2
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Hint: $X$ and $Y$ are independent iff the joint density is the product of the marginal densities.
$endgroup$
– Math1000
Dec 20 '18 at 3:33
add a comment |
2
$begingroup$
Hint: $X$ and $Y$ are independent iff the joint density is the product of the marginal densities.
$endgroup$
– Math1000
Dec 20 '18 at 3:33
2
2
$begingroup$
Hint: $X$ and $Y$ are independent iff the joint density is the product of the marginal densities.
$endgroup$
– Math1000
Dec 20 '18 at 3:33
$begingroup$
Hint: $X$ and $Y$ are independent iff the joint density is the product of the marginal densities.
$endgroup$
– Math1000
Dec 20 '18 at 3:33
add a comment |
1 Answer
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The key is what Math1000 says in the comments. We have $f_{XY}(x,y) = f_X(x)f_Y(y).$ A row of the table is for constant $x,$ so we have $$ f_{XY}(0, y) = f_X(0) f_Y(y)\ f_{XY}(1, y) = f_X(1) f_Y(y)$$ etc, so for instance $$ f_{XY}(0,y) = frac{f_X(1)}{f_X(0)}f_{XY}(1,y).$$
Some lingering questions:
- What if $f_X(i)=0$ for some or several $i$?
- This only shows one direction. How do you show that if the rows of the table are proportional, then $X$ and $Y$ are independent?
answered Dec 20 '18 at 5:42
spaceisdarkgreenspaceisdarkgreen
33.8k21753
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1 Answer
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1 Answer
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$begingroup$
The key is what Math1000 says in the comments. We have $f_{XY}(x,y) = f_X(x)f_Y(y).$ A row of the table is for constant $x,$ so we have $$ f_{XY}(0, y) = f_X(0) f_Y(y)\ f_{XY}(1, y) = f_X(1) f_Y(y)$$ etc, so for instance $$ f_{XY}(0,y) = frac{f_X(1)}{f_X(0)}f_{XY}(1,y).$$
Some lingering questions:
- What if $f_X(i)=0$ for some or several $i$?
- This only shows one direction. How do you show that if the rows of the table are proportional, then $X$ and $Y$ are independent?
answered Dec 20 '18 at 5:42
spaceisdarkgreenspaceisdarkgreen
33.8k21753
$endgroup$
add a comment |
$begingroup$
The key is what Math1000 says in the comments. We have $f_{XY}(x,y) = f_X(x)f_Y(y).$ A row of the table is for constant $x,$ so we have $$ f_{XY}(0, y) = f_X(0) f_Y(y)\ f_{XY}(1, y) = f_X(1) f_Y(y)$$ etc, so for instance $$ f_{XY}(0,y) = frac{f_X(1)}{f_X(0)}f_{XY}(1,y).$$
Some lingering questions:
- What if $f_X(i)=0$ for some or several $i$?
- This only shows one direction. How do you show that if the rows of the table are proportional, then $X$ and $Y$ are independent?
answered Dec 20 '18 at 5:42
spaceisdarkgreenspaceisdarkgreen
33.8k21753
$endgroup$
add a comment |
$begingroup$
The key is what Math1000 says in the comments. We have $f_{XY}(x,y) = f_X(x)f_Y(y).$ A row of the table is for constant $x,$ so we have $$ f_{XY}(0, y) = f_X(0) f_Y(y)\ f_{XY}(1, y) = f_X(1) f_Y(y)$$ etc, so for instance $$ f_{XY}(0,y) = frac{f_X(1)}{f_X(0)}f_{XY}(1,y).$$
Some lingering questions:
- What if $f_X(i)=0$ for some or several $i$?
- This only shows one direction. How do you show that if the rows of the table are proportional, then $X$ and $Y$ are independent?
answered Dec 20 '18 at 5:42
spaceisdarkgreenspaceisdarkgreen
33.8k21753
$endgroup$
The key is what Math1000 says in the comments. We have $f_{XY}(x,y) = f_X(x)f_Y(y).$ A row of the table is for constant $x,$ so we have $$ f_{XY}(0, y) = f_X(0) f_Y(y)\ f_{XY}(1, y) = f_X(1) f_Y(y)$$ etc, so for instance $$ f_{XY}(0,y) = frac{f_X(1)}{f_X(0)}f_{XY}(1,y).$$
Some lingering questions:
- What if $f_X(i)=0$ for some or several $i$?
- This only shows one direction. How do you show that if the rows of the table are proportional, then $X$ and $Y$ are independent?
answered Dec 20 '18 at 5:42
spaceisdarkgreenspaceisdarkgreen
33.8k21753
answered Dec 20 '18 at 5:42
spaceisdarkgreenspaceisdarkgreen
33.8k21753
answered Dec 20 '18 at 5:42
spaceisdarkgreenspaceisdarkgreen
33.8k21753
answered Dec 20 '18 at 5:42
spaceisdarkgreenspaceisdarkgreen
33.8k21753
33.8k21753
add a comment |
add a comment |
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$begingroup$
Hint: $X$ and $Y$ are independent iff the joint density is the product of the marginal densities.
$endgroup$
– Math1000
Dec 20 '18 at 3:33