Mixing
Sum of many Bernoulli random variable with different amplitude but same success probability
0
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Let $Y = a_1X_1 + a_2X_2 + a_3X_3 +....a_nX_n$, where $a_i $ are just constants and $X_i$ are independent Bernoulli random variables with probability of 0.5. Then what would be the distribution of Y? Thanks!
probability-distributions
asked Jan 30 at 9:07
Tianyu WangTianyu Wang
11
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add a comment |
0
$begingroup$
Let $Y = a_1X_1 + a_2X_2 + a_3X_3 +....a_nX_n$, where $a_i $ are just constants and $X_i$ are independent Bernoulli random variables with probability of 0.5. Then what would be the distribution of Y? Thanks!
probability-distributions
asked Jan 30 at 9:07
Tianyu WangTianyu Wang
11
$endgroup$
$begingroup$
What do you want to know exactly? Its CDF, PDF, characteristic function or if it's a known distribution?
$endgroup$
– Harnak
Jan 30 at 9:15
1
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If $n$ is large, you can approximate it with a Normal distribution.
$endgroup$
– Damien
Jan 30 at 9:17
$begingroup$
Harnak: I want to know the pmf, especially the derivative of pmf near the mean value of Y. Damien: is this because of the CLT?
$endgroup$
– Tianyu Wang
Jan 30 at 9:45
$begingroup$
Yes, because of the CLT. Note: when you address a comment to someone, don't forget to provide their name . @Harnak did not receive an alert about your comment to him
$endgroup$
– Damien
Jan 30 at 10:02
$begingroup$
You get a discrete distribution. Difficult to define a derivative of the pmf, except if you use a non discrete (Normal) approximation
$endgroup$
– Damien
Jan 30 at 10:10
add a comment |
0
0
0
$begingroup$
Let $Y = a_1X_1 + a_2X_2 + a_3X_3 +....a_nX_n$, where $a_i $ are just constants and $X_i$ are independent Bernoulli random variables with probability of 0.5. Then what would be the distribution of Y? Thanks!
probability-distributions
asked Jan 30 at 9:07
Tianyu WangTianyu Wang
11
$endgroup$
Let $Y = a_1X_1 + a_2X_2 + a_3X_3 +....a_nX_n$, where $a_i $ are just constants and $X_i$ are independent Bernoulli random variables with probability of 0.5. Then what would be the distribution of Y? Thanks!
probability-distributions
probability-distributions
asked Jan 30 at 9:07
Tianyu WangTianyu Wang
11
asked Jan 30 at 9:07
Tianyu WangTianyu Wang
11
asked Jan 30 at 9:07
Tianyu WangTianyu Wang
11
asked Jan 30 at 9:07
Tianyu WangTianyu Wang
11
asked Jan 30 at 9:07
Tianyu WangTianyu Wang
11
11
$begingroup$
What do you want to know exactly? Its CDF, PDF, characteristic function or if it's a known distribution?
$endgroup$
– Harnak
Jan 30 at 9:15
1
$begingroup$
If $n$ is large, you can approximate it with a Normal distribution.
$endgroup$
– Damien
Jan 30 at 9:17
$begingroup$
Harnak: I want to know the pmf, especially the derivative of pmf near the mean value of Y. Damien: is this because of the CLT?
$endgroup$
– Tianyu Wang
Jan 30 at 9:45
$begingroup$
Yes, because of the CLT. Note: when you address a comment to someone, don't forget to provide their name . @Harnak did not receive an alert about your comment to him
$endgroup$
– Damien
Jan 30 at 10:02
$begingroup$
You get a discrete distribution. Difficult to define a derivative of the pmf, except if you use a non discrete (Normal) approximation
$endgroup$
– Damien
Jan 30 at 10:10
add a comment |
$begingroup$
What do you want to know exactly? Its CDF, PDF, characteristic function or if it's a known distribution?
$endgroup$
– Harnak
Jan 30 at 9:15
1
$begingroup$
If $n$ is large, you can approximate it with a Normal distribution.
$endgroup$
– Damien
Jan 30 at 9:17
$begingroup$
Harnak: I want to know the pmf, especially the derivative of pmf near the mean value of Y. Damien: is this because of the CLT?
$endgroup$
– Tianyu Wang
Jan 30 at 9:45
$begingroup$
Yes, because of the CLT. Note: when you address a comment to someone, don't forget to provide their name . @Harnak did not receive an alert about your comment to him
$endgroup$
– Damien
Jan 30 at 10:02
$begingroup$
You get a discrete distribution. Difficult to define a derivative of the pmf, except if you use a non discrete (Normal) approximation
$endgroup$
– Damien
Jan 30 at 10:10
$begingroup$
What do you want to know exactly? Its CDF, PDF, characteristic function or if it's a known distribution?
$endgroup$
– Harnak
Jan 30 at 9:15
$begingroup$
What do you want to know exactly? Its CDF, PDF, characteristic function or if it's a known distribution?
$endgroup$
– Harnak
Jan 30 at 9:15
1
1
$begingroup$
If $n$ is large, you can approximate it with a Normal distribution.
$endgroup$
– Damien
Jan 30 at 9:17
$begingroup$
If $n$ is large, you can approximate it with a Normal distribution.
$endgroup$
– Damien
Jan 30 at 9:17
$begingroup$
Harnak: I want to know the pmf, especially the derivative of pmf near the mean value of Y. Damien: is this because of the CLT?
$endgroup$
– Tianyu Wang
Jan 30 at 9:45
$begingroup$
Harnak: I want to know the pmf, especially the derivative of pmf near the mean value of Y. Damien: is this because of the CLT?
$endgroup$
– Tianyu Wang
Jan 30 at 9:45
$begingroup$
Yes, because of the CLT. Note: when you address a comment to someone, don't forget to provide their name . @Harnak did not receive an alert about your comment to him
$endgroup$
– Damien
Jan 30 at 10:02
$begingroup$
Yes, because of the CLT. Note: when you address a comment to someone, don't forget to provide their name . @Harnak did not receive an alert about your comment to him
$endgroup$
– Damien
Jan 30 at 10:02
$begingroup$
You get a discrete distribution. Difficult to define a derivative of the pmf, except if you use a non discrete (Normal) approximation
$endgroup$
– Damien
Jan 30 at 10:10
$begingroup$
You get a discrete distribution. Difficult to define a derivative of the pmf, except if you use a non discrete (Normal) approximation
$endgroup$
– Damien
Jan 30 at 10:10
add a comment |
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$begingroup$
What do you want to know exactly? Its CDF, PDF, characteristic function or if it's a known distribution?
$endgroup$
– Harnak
Jan 30 at 9:15
1
$begingroup$
If $n$ is large, you can approximate it with a Normal distribution.
$endgroup$
– Damien
Jan 30 at 9:17
$begingroup$
Harnak: I want to know the pmf, especially the derivative of pmf near the mean value of Y. Damien: is this because of the CLT?
$endgroup$
– Tianyu Wang
Jan 30 at 9:45
$begingroup$
Yes, because of the CLT. Note: when you address a comment to someone, don't forget to provide their name . @Harnak did not receive an alert about your comment to him
$endgroup$
– Damien
Jan 30 at 10:02
$begingroup$
You get a discrete distribution. Difficult to define a derivative of the pmf, except if you use a non discrete (Normal) approximation
$endgroup$
– Damien
Jan 30 at 10:10