Best estimate for random values
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Due to work related issues I can't discuss the exact question I want to ask, but I thought of a silly little example that conveys the same idea.
Lets say the number of candy that comes in a package is a random variable with mean $mu$ and a standard deviation $s$, after about 2 months of data gathering we've got about 100000 measurements and a pretty good estimate of $mu$ and $s$.
Lets say that said candy comes in 5 flavours that are NOT identically distributed (we know the mean and standard deviation for each flavor, lets call them $mu_1$ through $mu_5$ and $s_1$ trough $s_5$).
Lets say that next month we will get a new batch (several packages) of candy from our supplier and we would like to estimate the amount of candy we will get for each flavour. Is there a better way than simply assuming that we'll get "around" the mean for each flavour taking into account that the amount of candy we'll get is around $mu$?
I have access to all the measurements made, so if anything is needed (higher order moments, other relevant data, etc.) I can compute it and update the question as needed.
Cheers and thanks!
statistics random estimation
asked Apr 2 '13 at 21:08
Zegpi
212
add a comment |Â
up vote
4
down vote
favorite
Due to work related issues I can't discuss the exact question I want to ask, but I thought of a silly little example that conveys the same idea.
Lets say the number of candy that comes in a package is a random variable with mean $mu$ and a standard deviation $s$, after about 2 months of data gathering we've got about 100000 measurements and a pretty good estimate of $mu$ and $s$.
Lets say that said candy comes in 5 flavours that are NOT identically distributed (we know the mean and standard deviation for each flavor, lets call them $mu_1$ through $mu_5$ and $s_1$ trough $s_5$).
Lets say that next month we will get a new batch (several packages) of candy from our supplier and we would like to estimate the amount of candy we will get for each flavour. Is there a better way than simply assuming that we'll get "around" the mean for each flavour taking into account that the amount of candy we'll get is around $mu$?
I have access to all the measurements made, so if anything is needed (higher order moments, other relevant data, etc.) I can compute it and update the question as needed.
Cheers and thanks!
statistics random estimation
asked Apr 2 '13 at 21:08
Zegpi
212
Is "batch" synonymous with "package"? If so, the question would be considerably clearer if you used the same word again. If not, please clarify the difference.
– joriki
Apr 2 '13 at 21:34
I'm sorry, I forgot to add that a batch is several packages, I don't think it matters as we'll evaluate per "package"
– Zegpi
Apr 2 '13 at 21:52
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Due to work related issues I can't discuss the exact question I want to ask, but I thought of a silly little example that conveys the same idea.
Lets say the number of candy that comes in a package is a random variable with mean $mu$ and a standard deviation $s$, after about 2 months of data gathering we've got about 100000 measurements and a pretty good estimate of $mu$ and $s$.
Lets say that said candy comes in 5 flavours that are NOT identically distributed (we know the mean and standard deviation for each flavor, lets call them $mu_1$ through $mu_5$ and $s_1$ trough $s_5$).
Lets say that next month we will get a new batch (several packages) of candy from our supplier and we would like to estimate the amount of candy we will get for each flavour. Is there a better way than simply assuming that we'll get "around" the mean for each flavour taking into account that the amount of candy we'll get is around $mu$?
I have access to all the measurements made, so if anything is needed (higher order moments, other relevant data, etc.) I can compute it and update the question as needed.
Cheers and thanks!
statistics random estimation
asked Apr 2 '13 at 21:08
Zegpi
212
Due to work related issues I can't discuss the exact question I want to ask, but I thought of a silly little example that conveys the same idea.
Lets say the number of candy that comes in a package is a random variable with mean $mu$ and a standard deviation $s$, after about 2 months of data gathering we've got about 100000 measurements and a pretty good estimate of $mu$ and $s$.
Lets say that said candy comes in 5 flavours that are NOT identically distributed (we know the mean and standard deviation for each flavor, lets call them $mu_1$ through $mu_5$ and $s_1$ trough $s_5$).
Lets say that next month we will get a new batch (several packages) of candy from our supplier and we would like to estimate the amount of candy we will get for each flavour. Is there a better way than simply assuming that we'll get "around" the mean for each flavour taking into account that the amount of candy we'll get is around $mu$?
I have access to all the measurements made, so if anything is needed (higher order moments, other relevant data, etc.) I can compute it and update the question as needed.
Cheers and thanks!
statistics random estimation
asked Apr 2 '13 at 21:08
Zegpi
212
edited Apr 2 '13 at 21:51
asked Apr 2 '13 at 21:08
Zegpi
212
asked Apr 2 '13 at 21:08
Zegpi
212
asked Apr 2 '13 at 21:08
Zegpi
212
212
Is "batch" synonymous with "package"? If so, the question would be considerably clearer if you used the same word again. If not, please clarify the difference.
– joriki
Apr 2 '13 at 21:34
I'm sorry, I forgot to add that a batch is several packages, I don't think it matters as we'll evaluate per "package"
– Zegpi
Apr 2 '13 at 21:52
add a comment |Â
Is "batch" synonymous with "package"? If so, the question would be considerably clearer if you used the same word again. If not, please clarify the difference.
– joriki
Apr 2 '13 at 21:34
I'm sorry, I forgot to add that a batch is several packages, I don't think it matters as we'll evaluate per "package"
– Zegpi
Apr 2 '13 at 21:52
Is "batch" synonymous with "package"? If so, the question would be considerably clearer if you used the same word again. If not, please clarify the difference.
– joriki
Apr 2 '13 at 21:34
Is "batch" synonymous with "package"? If so, the question would be considerably clearer if you used the same word again. If not, please clarify the difference.
– joriki
Apr 2 '13 at 21:34
I'm sorry, I forgot to add that a batch is several packages, I don't think it matters as we'll evaluate per "package"
– Zegpi
Apr 2 '13 at 21:52
I'm sorry, I forgot to add that a batch is several packages, I don't think it matters as we'll evaluate per "package"
– Zegpi
Apr 2 '13 at 21:52
add a comment |Â
1 Answer
1
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I find the question a bit unclear but since you have an estimate of the mean and standard deviation for a single package, assuming the candy in each package is independent why not use, $E(X_1+X_2)=E(X_1)+E(X_2)$ and $Var(X_1+X_2) = Var(X_1) + Var(X_2) + 2*cov(X_1,X_2)$ ?
answered Apr 2 '13 at 21:47
MrOperator
1886
Since we want to get an estimate for each "flavour" we are now using $mu_1$ through $mu_5$. I would like to know if there is a better method to estimate how many of each "flavour" we'll get.
– Zegpi
Apr 2 '13 at 22:00
I do no think that you can do beter then using the distirubtion of each flavour. If for example you would know that $sigma_1=sigma_2$ you could use a pooled variance estimator to get a better estimate $s_pooled$ for $sigma_1$ and $sigma_2$. The same reasoning holds for $mu$.
– MrOperator
Apr 3 '13 at 10:14
That's what I thought, but I hoped someone could have a better idea.
– Zegpi
Apr 3 '13 at 11:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I find the question a bit unclear but since you have an estimate of the mean and standard deviation for a single package, assuming the candy in each package is independent why not use, $E(X_1+X_2)=E(X_1)+E(X_2)$ and $Var(X_1+X_2) = Var(X_1) + Var(X_2) + 2*cov(X_1,X_2)$ ?
answered Apr 2 '13 at 21:47
MrOperator
1886
Since we want to get an estimate for each "flavour" we are now using $mu_1$ through $mu_5$. I would like to know if there is a better method to estimate how many of each "flavour" we'll get.
– Zegpi
Apr 2 '13 at 22:00
I do no think that you can do beter then using the distirubtion of each flavour. If for example you would know that $sigma_1=sigma_2$ you could use a pooled variance estimator to get a better estimate $s_pooled$ for $sigma_1$ and $sigma_2$. The same reasoning holds for $mu$.
– MrOperator
Apr 3 '13 at 10:14
That's what I thought, but I hoped someone could have a better idea.
– Zegpi
Apr 3 '13 at 11:46
add a comment |Â
up vote
0
down vote
I find the question a bit unclear but since you have an estimate of the mean and standard deviation for a single package, assuming the candy in each package is independent why not use, $E(X_1+X_2)=E(X_1)+E(X_2)$ and $Var(X_1+X_2) = Var(X_1) + Var(X_2) + 2*cov(X_1,X_2)$ ?
answered Apr 2 '13 at 21:47
MrOperator
1886
Since we want to get an estimate for each "flavour" we are now using $mu_1$ through $mu_5$. I would like to know if there is a better method to estimate how many of each "flavour" we'll get.
– Zegpi
Apr 2 '13 at 22:00
I do no think that you can do beter then using the distirubtion of each flavour. If for example you would know that $sigma_1=sigma_2$ you could use a pooled variance estimator to get a better estimate $s_pooled$ for $sigma_1$ and $sigma_2$. The same reasoning holds for $mu$.
– MrOperator
Apr 3 '13 at 10:14
That's what I thought, but I hoped someone could have a better idea.
– Zegpi
Apr 3 '13 at 11:46
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I find the question a bit unclear but since you have an estimate of the mean and standard deviation for a single package, assuming the candy in each package is independent why not use, $E(X_1+X_2)=E(X_1)+E(X_2)$ and $Var(X_1+X_2) = Var(X_1) + Var(X_2) + 2*cov(X_1,X_2)$ ?
answered Apr 2 '13 at 21:47
MrOperator
1886
I find the question a bit unclear but since you have an estimate of the mean and standard deviation for a single package, assuming the candy in each package is independent why not use, $E(X_1+X_2)=E(X_1)+E(X_2)$ and $Var(X_1+X_2) = Var(X_1) + Var(X_2) + 2*cov(X_1,X_2)$ ?
answered Apr 2 '13 at 21:47
MrOperator
1886
answered Apr 2 '13 at 21:47
MrOperator
1886
answered Apr 2 '13 at 21:47
MrOperator
1886
answered Apr 2 '13 at 21:47
MrOperator
1886
1886
Since we want to get an estimate for each "flavour" we are now using $mu_1$ through $mu_5$. I would like to know if there is a better method to estimate how many of each "flavour" we'll get.
– Zegpi
Apr 2 '13 at 22:00
I do no think that you can do beter then using the distirubtion of each flavour. If for example you would know that $sigma_1=sigma_2$ you could use a pooled variance estimator to get a better estimate $s_pooled$ for $sigma_1$ and $sigma_2$. The same reasoning holds for $mu$.
– MrOperator
Apr 3 '13 at 10:14
That's what I thought, but I hoped someone could have a better idea.
– Zegpi
Apr 3 '13 at 11:46
add a comment |Â
Since we want to get an estimate for each "flavour" we are now using $mu_1$ through $mu_5$. I would like to know if there is a better method to estimate how many of each "flavour" we'll get.
– Zegpi
Apr 2 '13 at 22:00
I do no think that you can do beter then using the distirubtion of each flavour. If for example you would know that $sigma_1=sigma_2$ you could use a pooled variance estimator to get a better estimate $s_pooled$ for $sigma_1$ and $sigma_2$. The same reasoning holds for $mu$.
– MrOperator
Apr 3 '13 at 10:14
That's what I thought, but I hoped someone could have a better idea.
– Zegpi
Apr 3 '13 at 11:46
Since we want to get an estimate for each "flavour" we are now using $mu_1$ through $mu_5$. I would like to know if there is a better method to estimate how many of each "flavour" we'll get.
– Zegpi
Apr 2 '13 at 22:00
Since we want to get an estimate for each "flavour" we are now using $mu_1$ through $mu_5$. I would like to know if there is a better method to estimate how many of each "flavour" we'll get.
– Zegpi
Apr 2 '13 at 22:00
I do no think that you can do beter then using the distirubtion of each flavour. If for example you would know that $sigma_1=sigma_2$ you could use a pooled variance estimator to get a better estimate $s_pooled$ for $sigma_1$ and $sigma_2$. The same reasoning holds for $mu$.
– MrOperator
Apr 3 '13 at 10:14
I do no think that you can do beter then using the distirubtion of each flavour. If for example you would know that $sigma_1=sigma_2$ you could use a pooled variance estimator to get a better estimate $s_pooled$ for $sigma_1$ and $sigma_2$. The same reasoning holds for $mu$.
– MrOperator
Apr 3 '13 at 10:14
That's what I thought, but I hoped someone could have a better idea.
– Zegpi
Apr 3 '13 at 11:46
That's what I thought, but I hoped someone could have a better idea.
– Zegpi
Apr 3 '13 at 11:46
add a comment |Â
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Is "batch" synonymous with "package"? If so, the question would be considerably clearer if you used the same word again. If not, please clarify the difference.
– joriki
Apr 2 '13 at 21:34
I'm sorry, I forgot to add that a batch is several packages, I don't think it matters as we'll evaluate per "package"
– Zegpi
Apr 2 '13 at 21:52