Probability density of a random variable [closed]
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How to calculate the probability density of a random variable being the sum of independent random variables $X _{1}$ and $X _{2}$ with normal distributions $N( m _{1}, partial ^{2} _{1} )$ and $N( m _{2}, partial ^{2} _{2} )$?
Any help?
probability
asked Dec 28 '18 at 9:17
AbakusAbakus
32
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closed as off-topic by Kavi Rama Murthy, StubbornAtom, NCh, Eevee Trainer, Andrew Dec 29 '18 at 0:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, StubbornAtom, NCh, Eevee Trainer, Andrew
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How to calculate the probability density of a random variable being the sum of independent random variables $X _{1}$ and $X _{2}$ with normal distributions $N( m _{1}, partial ^{2} _{1} )$ and $N( m _{2}, partial ^{2} _{2} )$?
Any help?
probability
asked Dec 28 '18 at 9:17
AbakusAbakus
32
$endgroup$
closed as off-topic by Kavi Rama Murthy, StubbornAtom, NCh, Eevee Trainer, Andrew Dec 29 '18 at 0:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, StubbornAtom, NCh, Eevee Trainer, Andrew
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
This is one of those questions that is so readily answered in textbooks—even a google search of the phrase “sums of normal random variables” leads one directly to the wikipedia page titled so which explicitly answers the question—that it is extremely apparent you have not put sufficient thought or effort into the question yourself. This is not a free homework website and is not advertised as such...
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– LoveTooNap29
Dec 28 '18 at 16:48
add a comment |
$begingroup$
How to calculate the probability density of a random variable being the sum of independent random variables $X _{1}$ and $X _{2}$ with normal distributions $N( m _{1}, partial ^{2} _{1} )$ and $N( m _{2}, partial ^{2} _{2} )$?
Any help?
probability
asked Dec 28 '18 at 9:17
AbakusAbakus
32
$endgroup$
How to calculate the probability density of a random variable being the sum of independent random variables $X _{1}$ and $X _{2}$ with normal distributions $N( m _{1}, partial ^{2} _{1} )$ and $N( m _{2}, partial ^{2} _{2} )$?
Any help?
probability
probability
asked Dec 28 '18 at 9:17
AbakusAbakus
32
asked Dec 28 '18 at 9:17
AbakusAbakus
32
asked Dec 28 '18 at 9:17
AbakusAbakus
32
asked Dec 28 '18 at 9:17
AbakusAbakus
32
asked Dec 28 '18 at 9:17
AbakusAbakus
32
32
closed as off-topic by Kavi Rama Murthy, StubbornAtom, NCh, Eevee Trainer, Andrew Dec 29 '18 at 0:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, StubbornAtom, NCh, Eevee Trainer, Andrew
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Kavi Rama Murthy, StubbornAtom, NCh, Eevee Trainer, Andrew Dec 29 '18 at 0:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, StubbornAtom, NCh, Eevee Trainer, Andrew
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
This is one of those questions that is so readily answered in textbooks—even a google search of the phrase “sums of normal random variables” leads one directly to the wikipedia page titled so which explicitly answers the question—that it is extremely apparent you have not put sufficient thought or effort into the question yourself. This is not a free homework website and is not advertised as such...
$endgroup$
– LoveTooNap29
Dec 28 '18 at 16:48
add a comment |
$begingroup$
This is one of those questions that is so readily answered in textbooks—even a google search of the phrase “sums of normal random variables” leads one directly to the wikipedia page titled so which explicitly answers the question—that it is extremely apparent you have not put sufficient thought or effort into the question yourself. This is not a free homework website and is not advertised as such...
$endgroup$
– LoveTooNap29
Dec 28 '18 at 16:48
$begingroup$
This is one of those questions that is so readily answered in textbooks—even a google search of the phrase “sums of normal random variables” leads one directly to the wikipedia page titled so which explicitly answers the question—that it is extremely apparent you have not put sufficient thought or effort into the question yourself. This is not a free homework website and is not advertised as such...
$endgroup$
– LoveTooNap29
Dec 28 '18 at 16:48
$begingroup$
This is one of those questions that is so readily answered in textbooks—even a google search of the phrase “sums of normal random variables” leads one directly to the wikipedia page titled so which explicitly answers the question—that it is extremely apparent you have not put sufficient thought or effort into the question yourself. This is not a free homework website and is not advertised as such...
$endgroup$
– LoveTooNap29
Dec 28 '18 at 16:48
add a comment |
1 Answer
1
active
oldest
votes
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The sum of two normal random variables is normal (see https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables) from there you can calculate the mean and variance of the distributions: $mu = m_1 + m_2$ and same for the variance.
answered Dec 28 '18 at 9:26
Robin NicoleRobin Nicole
20112
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“and same for the variance”—the variance of a sum of normal RVs is only the sum of variances in the case of independence.
$endgroup$
– LoveTooNap29
Dec 28 '18 at 18:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The sum of two normal random variables is normal (see https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables) from there you can calculate the mean and variance of the distributions: $mu = m_1 + m_2$ and same for the variance.
answered Dec 28 '18 at 9:26
Robin NicoleRobin Nicole
20112
$endgroup$
$begingroup$
“and same for the variance”—the variance of a sum of normal RVs is only the sum of variances in the case of independence.
$endgroup$
– LoveTooNap29
Dec 28 '18 at 18:42
add a comment |
$begingroup$
The sum of two normal random variables is normal (see https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables) from there you can calculate the mean and variance of the distributions: $mu = m_1 + m_2$ and same for the variance.
answered Dec 28 '18 at 9:26
Robin NicoleRobin Nicole
20112
$endgroup$
$begingroup$
“and same for the variance”—the variance of a sum of normal RVs is only the sum of variances in the case of independence.
$endgroup$
– LoveTooNap29
Dec 28 '18 at 18:42
add a comment |
$begingroup$
The sum of two normal random variables is normal (see https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables) from there you can calculate the mean and variance of the distributions: $mu = m_1 + m_2$ and same for the variance.
answered Dec 28 '18 at 9:26
Robin NicoleRobin Nicole
20112
$endgroup$
The sum of two normal random variables is normal (see https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables) from there you can calculate the mean and variance of the distributions: $mu = m_1 + m_2$ and same for the variance.
answered Dec 28 '18 at 9:26
Robin NicoleRobin Nicole
20112
answered Dec 28 '18 at 9:26
Robin NicoleRobin Nicole
20112
answered Dec 28 '18 at 9:26
Robin NicoleRobin Nicole
20112
answered Dec 28 '18 at 9:26
Robin NicoleRobin Nicole
20112
20112
$begingroup$
“and same for the variance”—the variance of a sum of normal RVs is only the sum of variances in the case of independence.
$endgroup$
– LoveTooNap29
Dec 28 '18 at 18:42
add a comment |
$begingroup$
“and same for the variance”—the variance of a sum of normal RVs is only the sum of variances in the case of independence.
$endgroup$
– LoveTooNap29
Dec 28 '18 at 18:42
$begingroup$
“and same for the variance”—the variance of a sum of normal RVs is only the sum of variances in the case of independence.
$endgroup$
– LoveTooNap29
Dec 28 '18 at 18:42
$begingroup$
“and same for the variance”—the variance of a sum of normal RVs is only the sum of variances in the case of independence.
$endgroup$
– LoveTooNap29
Dec 28 '18 at 18:42
add a comment |
$begingroup$
This is one of those questions that is so readily answered in textbooks—even a google search of the phrase “sums of normal random variables” leads one directly to the wikipedia page titled so which explicitly answers the question—that it is extremely apparent you have not put sufficient thought or effort into the question yourself. This is not a free homework website and is not advertised as such...
$endgroup$
– LoveTooNap29
Dec 28 '18 at 16:48