Probability Distribution verification
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a probability distribution must satisfy two conditions
1- the probability of each value of the random variable is between 0 and 1.
2- the sum over all the probabilities is equal to 1.
I think to exclude all the zero's mass probability, I should take the summation over all the support of x,y,z.
Can anyone please help me in how to write the verification. I know the concept but donot know how to write it.
Thanks
probability probability-distributions
asked Jan 14 at 1:54
Salwa MostafaSalwa Mostafa
274
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add a comment |
$begingroup$
a probability distribution must satisfy two conditions
1- the probability of each value of the random variable is between 0 and 1.
2- the sum over all the probabilities is equal to 1.
I think to exclude all the zero's mass probability, I should take the summation over all the support of x,y,z.
Can anyone please help me in how to write the verification. I know the concept but donot know how to write it.
Thanks
probability probability-distributions
asked Jan 14 at 1:54
Salwa MostafaSalwa Mostafa
274
$endgroup$
$begingroup$
Are you trying to write software? What does it mean to “write the verification”?
$endgroup$
– NicNic8
Jan 14 at 2:40
$begingroup$
I mean the answer to the question as a proof
$endgroup$
– Salwa Mostafa
Jan 14 at 2:52
add a comment |
$begingroup$
a probability distribution must satisfy two conditions
1- the probability of each value of the random variable is between 0 and 1.
2- the sum over all the probabilities is equal to 1.
I think to exclude all the zero's mass probability, I should take the summation over all the support of x,y,z.
Can anyone please help me in how to write the verification. I know the concept but donot know how to write it.
Thanks
probability probability-distributions
asked Jan 14 at 1:54
Salwa MostafaSalwa Mostafa
274
$endgroup$
a probability distribution must satisfy two conditions
1- the probability of each value of the random variable is between 0 and 1.
2- the sum over all the probabilities is equal to 1.
I think to exclude all the zero's mass probability, I should take the summation over all the support of x,y,z.
Can anyone please help me in how to write the verification. I know the concept but donot know how to write it.
Thanks
probability probability-distributions
probability probability-distributions
asked Jan 14 at 1:54
Salwa MostafaSalwa Mostafa
274
asked Jan 14 at 1:54
Salwa MostafaSalwa Mostafa
274
asked Jan 14 at 1:54
Salwa MostafaSalwa Mostafa
274
asked Jan 14 at 1:54
Salwa MostafaSalwa Mostafa
274
asked Jan 14 at 1:54
Salwa MostafaSalwa Mostafa
274
274
$begingroup$
Are you trying to write software? What does it mean to “write the verification”?
$endgroup$
– NicNic8
Jan 14 at 2:40
$begingroup$
I mean the answer to the question as a proof
$endgroup$
– Salwa Mostafa
Jan 14 at 2:52
add a comment |
$begingroup$
Are you trying to write software? What does it mean to “write the verification”?
$endgroup$
– NicNic8
Jan 14 at 2:40
$begingroup$
I mean the answer to the question as a proof
$endgroup$
– Salwa Mostafa
Jan 14 at 2:52
$begingroup$
Are you trying to write software? What does it mean to “write the verification”?
$endgroup$
– NicNic8
Jan 14 at 2:40
$begingroup$
Are you trying to write software? What does it mean to “write the verification”?
$endgroup$
– NicNic8
Jan 14 at 2:40
$begingroup$
I mean the answer to the question as a proof
$endgroup$
– Salwa Mostafa
Jan 14 at 2:52
$begingroup$
I mean the answer to the question as a proof
$endgroup$
– Salwa Mostafa
Jan 14 at 2:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $p(x,y,z) geq 0$ by definition the only thing you have to prove is that $sum_{x,y,z} p(x,y,z)=1$. First take the sum over $x$. Since $sum_x frac {p(x,y)} {p(y)}=1$ we are left with the proof of $sum_{y,z} p(y,z)=1$ which is clear.
answered Jan 14 at 6:05
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
$endgroup$
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
$begingroup$
Since $p(x,y,z) geq 0$ by definition the only thing you have to prove is that $sum_{x,y,z} p(x,y,z)=1$. First take the sum over $x$. Since $sum_x frac {p(x,y)} {p(y)}=1$ we are left with the proof of $sum_{y,z} p(y,z)=1$ which is clear.
answered Jan 14 at 6:05
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
$endgroup$
add a comment |
$begingroup$
Since $p(x,y,z) geq 0$ by definition the only thing you have to prove is that $sum_{x,y,z} p(x,y,z)=1$. First take the sum over $x$. Since $sum_x frac {p(x,y)} {p(y)}=1$ we are left with the proof of $sum_{y,z} p(y,z)=1$ which is clear.
answered Jan 14 at 6:05
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
$endgroup$
add a comment |
$begingroup$
Since $p(x,y,z) geq 0$ by definition the only thing you have to prove is that $sum_{x,y,z} p(x,y,z)=1$. First take the sum over $x$. Since $sum_x frac {p(x,y)} {p(y)}=1$ we are left with the proof of $sum_{y,z} p(y,z)=1$ which is clear.
answered Jan 14 at 6:05
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
$endgroup$
Since $p(x,y,z) geq 0$ by definition the only thing you have to prove is that $sum_{x,y,z} p(x,y,z)=1$. First take the sum over $x$. Since $sum_x frac {p(x,y)} {p(y)}=1$ we are left with the proof of $sum_{y,z} p(y,z)=1$ which is clear.
answered Jan 14 at 6:05
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
answered Jan 14 at 6:05
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
answered Jan 14 at 6:05
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
answered Jan 14 at 6:05
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
68.9k53169
add a comment |
add a comment |
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$begingroup$
Are you trying to write software? What does it mean to “write the verification”?
$endgroup$
– NicNic8
Jan 14 at 2:40
$begingroup$
I mean the answer to the question as a proof
$endgroup$
– Salwa Mostafa
Jan 14 at 2:52