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PMF conditioned on an Event
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OK, so i cannot understand how to derive relationship between unconditional and conditional PMF, given an Event A, with P(A) > 0.
I know that $p_X(x) = P(A) times p_A(x)$, but from definition of conditional probability, $p_A(x) = fracP(X=xcap A)P(A)$
Also by definition $p_X(x) = P(X=x)$. Combining these I get that $P(X=x) = P(X=xcap A)$. Where am I going wrong?
probability discrete-mathematics
asked Jul 23 at 19:46
Eugene
11
add a comment |Â
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OK, so i cannot understand how to derive relationship between unconditional and conditional PMF, given an Event A, with P(A) > 0.
I know that $p_X(x) = P(A) times p_A(x)$, but from definition of conditional probability, $p_A(x) = fracP(X=xcap A)P(A)$
Also by definition $p_X(x) = P(X=x)$. Combining these I get that $P(X=x) = P(X=xcap A)$. Where am I going wrong?
probability discrete-mathematics
asked Jul 23 at 19:46
Eugene
11
It is not true that $p_X(x) = P(A)times p_Xmid A(x)$.
– Math1000
Jul 23 at 22:17
It works on any examples that I have tried, and also I've seen this in a few youtube and online tutorials. However I would like to see explanation/derivation from a more reliable source.
– Eugene
Jul 24 at 7:35
If $p_X(x)=P(A)times p_Xmid A(x)$ then $p_Xmid A(x) = p_X(x)/P(A)$. Since $p_Xmid A(x) = mathbb P(X=xcap A)/P(A)$, it follows that $$p_X(x) = mathbb P(X=xcap A), $$ which clearly is not true in general.
– Math1000
Jul 24 at 8:16
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
OK, so i cannot understand how to derive relationship between unconditional and conditional PMF, given an Event A, with P(A) > 0.
I know that $p_X(x) = P(A) times p_A(x)$, but from definition of conditional probability, $p_A(x) = fracP(X=xcap A)P(A)$
Also by definition $p_X(x) = P(X=x)$. Combining these I get that $P(X=x) = P(X=xcap A)$. Where am I going wrong?
probability discrete-mathematics
asked Jul 23 at 19:46
Eugene
11
OK, so i cannot understand how to derive relationship between unconditional and conditional PMF, given an Event A, with P(A) > 0.
I know that $p_X(x) = P(A) times p_A(x)$, but from definition of conditional probability, $p_A(x) = fracP(X=xcap A)P(A)$
Also by definition $p_X(x) = P(X=x)$. Combining these I get that $P(X=x) = P(X=xcap A)$. Where am I going wrong?
probability discrete-mathematics
asked Jul 23 at 19:46
Eugene
11
asked Jul 23 at 19:46
Eugene
11
asked Jul 23 at 19:46
Eugene
11
asked Jul 23 at 19:46
Eugene
11
11
It is not true that $p_X(x) = P(A)times p_Xmid A(x)$.
– Math1000
Jul 23 at 22:17
It works on any examples that I have tried, and also I've seen this in a few youtube and online tutorials. However I would like to see explanation/derivation from a more reliable source.
– Eugene
Jul 24 at 7:35
If $p_X(x)=P(A)times p_Xmid A(x)$ then $p_Xmid A(x) = p_X(x)/P(A)$. Since $p_Xmid A(x) = mathbb P(X=xcap A)/P(A)$, it follows that $$p_X(x) = mathbb P(X=xcap A), $$ which clearly is not true in general.
– Math1000
Jul 24 at 8:16
add a comment |Â
It is not true that $p_X(x) = P(A)times p_Xmid A(x)$.
– Math1000
Jul 23 at 22:17
It works on any examples that I have tried, and also I've seen this in a few youtube and online tutorials. However I would like to see explanation/derivation from a more reliable source.
– Eugene
Jul 24 at 7:35
If $p_X(x)=P(A)times p_Xmid A(x)$ then $p_Xmid A(x) = p_X(x)/P(A)$. Since $p_Xmid A(x) = mathbb P(X=xcap A)/P(A)$, it follows that $$p_X(x) = mathbb P(X=xcap A), $$ which clearly is not true in general.
– Math1000
Jul 24 at 8:16
It is not true that $p_X(x) = P(A)times p_Xmid A(x)$.
– Math1000
Jul 23 at 22:17
It is not true that $p_X(x) = P(A)times p_Xmid A(x)$.
– Math1000
Jul 23 at 22:17
It works on any examples that I have tried, and also I've seen this in a few youtube and online tutorials. However I would like to see explanation/derivation from a more reliable source.
– Eugene
Jul 24 at 7:35
It works on any examples that I have tried, and also I've seen this in a few youtube and online tutorials. However I would like to see explanation/derivation from a more reliable source.
– Eugene
Jul 24 at 7:35
If $p_X(x)=P(A)times p_Xmid A(x)$ then $p_Xmid A(x) = p_X(x)/P(A)$. Since $p_Xmid A(x) = mathbb P(X=xcap A)/P(A)$, it follows that $$p_X(x) = mathbb P(X=xcap A), $$ which clearly is not true in general.
– Math1000
Jul 24 at 8:16
If $p_X(x)=P(A)times p_Xmid A(x)$ then $p_Xmid A(x) = p_X(x)/P(A)$. Since $p_Xmid A(x) = mathbb P(X=xcap A)/P(A)$, it follows that $$p_X(x) = mathbb P(X=xcap A), $$ which clearly is not true in general.
– Math1000
Jul 24 at 8:16
add a comment |Â
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It is not true that $p_X(x) = P(A)times p_Xmid A(x)$.
– Math1000
Jul 23 at 22:17
It works on any examples that I have tried, and also I've seen this in a few youtube and online tutorials. However I would like to see explanation/derivation from a more reliable source.
– Eugene
Jul 24 at 7:35
If $p_X(x)=P(A)times p_Xmid A(x)$ then $p_Xmid A(x) = p_X(x)/P(A)$. Since $p_Xmid A(x) = mathbb P(X=xcap A)/P(A)$, it follows that $$p_X(x) = mathbb P(X=xcap A), $$ which clearly is not true in general.
– Math1000
Jul 24 at 8:16