Mixing
writing probability function mathematically (in the shape of $f(x)$}…}
$begingroup$
I am trying to write the following expression mathematically: let $X_1,X_2$ be independent random variables, each having the probability function $P_X(k)=p(1-p)^k$ for k=0,1,... and $0<p<1$.
Is it something like $forall X_1X_2 exists p_X(k)=p(1-p)^k|(k=0,1,2,..) land (0<p<1)$ ?
I would appreciate a correction and explanation, so I could not repeat the same mistake in the future.
Note: I did not define what is an independent random variable there; I just gave a general expression for random variables $X_1,X_2$. If you could, I would appreciate seeing how to write it with the definition of independence.
Thank you very much!
probability-theory logic conditional-probability
edited Jan 28 at 16:07
J. W. Tanner
4,0171320
asked Jan 28 at 14:34
q123q123
75
$endgroup$
add a comment |
$begingroup$
I am trying to write the following expression mathematically: let $X_1,X_2$ be independent random variables, each having the probability function $P_X(k)=p(1-p)^k$ for k=0,1,... and $0<p<1$.
Is it something like $forall X_1X_2 exists p_X(k)=p(1-p)^k|(k=0,1,2,..) land (0<p<1)$ ?
I would appreciate a correction and explanation, so I could not repeat the same mistake in the future.
Note: I did not define what is an independent random variable there; I just gave a general expression for random variables $X_1,X_2$. If you could, I would appreciate seeing how to write it with the definition of independence.
Thank you very much!
probability-theory logic conditional-probability
edited Jan 28 at 16:07
J. W. Tanner
4,0171320
asked Jan 28 at 14:34
q123q123
75
$endgroup$
$begingroup$
$X_1$ and $X_2$ are "fixed"; if so, we cannot quantify them.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
1
$begingroup$
$P_X(k)$ must be (I think) $P_{X_i}(k)$ for $i=1,2$.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
add a comment |
$begingroup$
I am trying to write the following expression mathematically: let $X_1,X_2$ be independent random variables, each having the probability function $P_X(k)=p(1-p)^k$ for k=0,1,... and $0<p<1$.
Is it something like $forall X_1X_2 exists p_X(k)=p(1-p)^k|(k=0,1,2,..) land (0<p<1)$ ?
I would appreciate a correction and explanation, so I could not repeat the same mistake in the future.
Note: I did not define what is an independent random variable there; I just gave a general expression for random variables $X_1,X_2$. If you could, I would appreciate seeing how to write it with the definition of independence.
Thank you very much!
probability-theory logic conditional-probability
edited Jan 28 at 16:07
J. W. Tanner
4,0171320
asked Jan 28 at 14:34
q123q123
75
$endgroup$
I am trying to write the following expression mathematically: let $X_1,X_2$ be independent random variables, each having the probability function $P_X(k)=p(1-p)^k$ for k=0,1,... and $0<p<1$.
Is it something like $forall X_1X_2 exists p_X(k)=p(1-p)^k|(k=0,1,2,..) land (0<p<1)$ ?
I would appreciate a correction and explanation, so I could not repeat the same mistake in the future.
Note: I did not define what is an independent random variable there; I just gave a general expression for random variables $X_1,X_2$. If you could, I would appreciate seeing how to write it with the definition of independence.
Thank you very much!
probability-theory logic conditional-probability
probability-theory logic conditional-probability
edited Jan 28 at 16:07
J. W. Tanner
4,0171320
asked Jan 28 at 14:34
q123q123
75
edited Jan 28 at 16:07
J. W. Tanner
4,0171320
asked Jan 28 at 14:34
q123q123
75
edited Jan 28 at 16:07
J. W. Tanner
4,0171320
edited Jan 28 at 16:07
J. W. Tanner
4,0171320
edited Jan 28 at 16:07
J. W. Tanner
4,0171320
4,0171320
asked Jan 28 at 14:34
q123q123
75
asked Jan 28 at 14:34
q123q123
75
asked Jan 28 at 14:34
q123q123
75
75
$begingroup$
$X_1$ and $X_2$ are "fixed"; if so, we cannot quantify them.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
1
$begingroup$
$P_X(k)$ must be (I think) $P_{X_i}(k)$ for $i=1,2$.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
add a comment |
$begingroup$
$X_1$ and $X_2$ are "fixed"; if so, we cannot quantify them.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
1
$begingroup$
$P_X(k)$ must be (I think) $P_{X_i}(k)$ for $i=1,2$.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
$begingroup$
$X_1$ and $X_2$ are "fixed"; if so, we cannot quantify them.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
$begingroup$
$X_1$ and $X_2$ are "fixed"; if so, we cannot quantify them.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
1
1
$begingroup$
$P_X(k)$ must be (I think) $P_{X_i}(k)$ for $i=1,2$.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
$begingroup$
$P_X(k)$ must be (I think) $P_{X_i}(k)$ for $i=1,2$.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
add a comment |
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$begingroup$
$X_1$ and $X_2$ are "fixed"; if so, we cannot quantify them.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40
1
$begingroup$
$P_X(k)$ must be (I think) $P_{X_i}(k)$ for $i=1,2$.
$endgroup$
– Mauro ALLEGRANZA
Jan 28 at 14:40