marginal probabilities, multivariate random variables
$begingroup$
I want to solve the task below...
However, I have a problem with the marginal probabilities not adding up to 1.
what's wrong?
probability probability-theory marginal-probability
asked Dec 8 '18 at 14:19
thebillythebilly
566
$endgroup$
add a comment |
$begingroup$
I want to solve the task below...
However, I have a problem with the marginal probabilities not adding up to 1.
what's wrong?
probability probability-theory marginal-probability
asked Dec 8 '18 at 14:19
thebillythebilly
566
$endgroup$
add a comment |
$begingroup$
I want to solve the task below...
However, I have a problem with the marginal probabilities not adding up to 1.
what's wrong?
probability probability-theory marginal-probability
asked Dec 8 '18 at 14:19
thebillythebilly
566
$endgroup$
I want to solve the task below...
However, I have a problem with the marginal probabilities not adding up to 1.
what's wrong?
probability probability-theory marginal-probability
probability probability-theory marginal-probability
asked Dec 8 '18 at 14:19
thebillythebilly
566
asked Dec 8 '18 at 14:19
thebillythebilly
566
asked Dec 8 '18 at 14:19
thebillythebilly
566
asked Dec 8 '18 at 14:19
thebillythebilly
566
asked Dec 8 '18 at 14:19
thebillythebilly
566
566
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Order matters.
For $(0,1)$, it should be $2 cdot frac36 cdot frac16$ as we can switch the order.
Similarly, probability for $(1,0)$ and $(1,1)$ needs to be multiplied by $2$.
answered Dec 8 '18 at 14:33
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Ah, I See. Thank you.
$endgroup$
– thebilly
Dec 8 '18 at 17:34
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Order matters.
For $(0,1)$, it should be $2 cdot frac36 cdot frac16$ as we can switch the order.
Similarly, probability for $(1,0)$ and $(1,1)$ needs to be multiplied by $2$.
answered Dec 8 '18 at 14:33
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Ah, I See. Thank you.
$endgroup$
– thebilly
Dec 8 '18 at 17:34
add a comment |
$begingroup$
Order matters.
For $(0,1)$, it should be $2 cdot frac36 cdot frac16$ as we can switch the order.
Similarly, probability for $(1,0)$ and $(1,1)$ needs to be multiplied by $2$.
answered Dec 8 '18 at 14:33
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Ah, I See. Thank you.
$endgroup$
– thebilly
Dec 8 '18 at 17:34
add a comment |
$begingroup$
Order matters.
For $(0,1)$, it should be $2 cdot frac36 cdot frac16$ as we can switch the order.
Similarly, probability for $(1,0)$ and $(1,1)$ needs to be multiplied by $2$.
answered Dec 8 '18 at 14:33
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
Order matters.
For $(0,1)$, it should be $2 cdot frac36 cdot frac16$ as we can switch the order.
Similarly, probability for $(1,0)$ and $(1,1)$ needs to be multiplied by $2$.
answered Dec 8 '18 at 14:33
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 8 '18 at 14:33
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 8 '18 at 14:33
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 8 '18 at 14:33
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
$begingroup$
Ah, I See. Thank you.
$endgroup$
– thebilly
Dec 8 '18 at 17:34
add a comment |
$begingroup$
Ah, I See. Thank you.
$endgroup$
– thebilly
Dec 8 '18 at 17:34
$begingroup$
Ah, I See. Thank you.
$endgroup$
– thebilly
Dec 8 '18 at 17:34
$begingroup$
Ah, I See. Thank you.
$endgroup$
– thebilly
Dec 8 '18 at 17:34
add a comment |
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