Sampling from Gaussian mixture models, when are the sampled data independent?












2












$begingroup$


Suppose I generate a Gaussian mixture model with $N$ Gaussian distributions



$p(x) = sumlimits_{i = 1}^N w_i mathcal{N}(x;mu_i, Sigma_i)$



where $w_i$ are the weights.



Now I sample some points ${x_n}$ from $p(x)$



What condition is needed to ensure that the points are independent from each other?



Is it sufficient to assume that each component $mathcal{N}(x;mu_i, Sigma_i)$ have diagonal covariance matrix $Sigma_i$?






normal-distribution mixed-model random-variable independence gaussian-mixture







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$endgroup$

















    2












    $begingroup$


    Suppose I generate a Gaussian mixture model with $N$ Gaussian distributions



    $p(x) = sumlimits_{i = 1}^N w_i mathcal{N}(x;mu_i, Sigma_i)$



    where $w_i$ are the weights.



    Now I sample some points ${x_n}$ from $p(x)$



    What condition is needed to ensure that the points are independent from each other?



    Is it sufficient to assume that each component $mathcal{N}(x;mu_i, Sigma_i)$ have diagonal covariance matrix $Sigma_i$?






    normal-distribution mixed-model random-variable independence gaussian-mixture







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Suppose I generate a Gaussian mixture model with $N$ Gaussian distributions



      $p(x) = sumlimits_{i = 1}^N w_i mathcal{N}(x;mu_i, Sigma_i)$



      where $w_i$ are the weights.



      Now I sample some points ${x_n}$ from $p(x)$



      What condition is needed to ensure that the points are independent from each other?



      Is it sufficient to assume that each component $mathcal{N}(x;mu_i, Sigma_i)$ have diagonal covariance matrix $Sigma_i$?






      normal-distribution mixed-model random-variable independence gaussian-mixture







      share|cite|improve this question









      $endgroup$




      Suppose I generate a Gaussian mixture model with $N$ Gaussian distributions



      $p(x) = sumlimits_{i = 1}^N w_i mathcal{N}(x;mu_i, Sigma_i)$



      where $w_i$ are the weights.



      Now I sample some points ${x_n}$ from $p(x)$



      What condition is needed to ensure that the points are independent from each other?



      Is it sufficient to assume that each component $mathcal{N}(x;mu_i, Sigma_i)$ have diagonal covariance matrix $Sigma_i$?






      normal-distribution mixed-model random-variable independence gaussian-mixture




      normal-distribution mixed-model random-variable independence gaussian-mixture






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 7 hours ago









      Shamisen ExpertShamisen Expert

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          $begingroup$

          You are confusing independence between the simulated vectors and independence between the components of a single vector. Unless otherwise stated, two simulations $X_1,X_2$ from$$sumlimits_{i = 1}^N w_i mathcal{N}(x;mu_i, Sigma_i)tag{1}$$will be independent, while the components of the vector $X_1$ will most likely be dependent, even when the matrices $Sigma_i$ are diagonal. (The intuition about that last point is that if $X_{11}$ is closer to $mu_{1k}$ than to the other Normal means, then the other components of $X_1$ are also more likely to originate from this same $k$-th element of the mixture.) For instance a most standard way to simulate from (1) is to simulate
          $$Z_1simmathcal M(1;w_1,ldots,w_N)qquad X_1|Z_1=k sim mathcal{N}(x;mu_k, Sigma_k)$$and$$Z_2simmathcal M(1;w_1,ldots,w_N)qquad X_2|Z_2=k sim mathcal{N}(x;mu_k, Sigma_k)$$If the four variables are mutually independent then $X_1$ and $X_2$ are independent.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. What is the meaning of $mathcal{M}$? Can you justify intuitively why the $X_1, X_2$ will most likely to be independent (this is clearer to me), whereas the components of $X_1$ will be dependent (this is not as clear to me)?
            $endgroup$
            – Shamisen Expert
            7 hours ago










          • $begingroup$
            Xi'an is using $mathcal{M}$ to denote the multinomial distribution. With the first parameter $n$ set to 1, the multinomial is also known as the categorical distribution. A categorical random variable takes on one of a finite set of categorical values with probability determined by the second parameter, the $w$ vector. In other words, $Z_i$ represents the group membership of observation $i$.
            $endgroup$
            – mb7744
            59 mins ago













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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          You are confusing independence between the simulated vectors and independence between the components of a single vector. Unless otherwise stated, two simulations $X_1,X_2$ from$$sumlimits_{i = 1}^N w_i mathcal{N}(x;mu_i, Sigma_i)tag{1}$$will be independent, while the components of the vector $X_1$ will most likely be dependent, even when the matrices $Sigma_i$ are diagonal. (The intuition about that last point is that if $X_{11}$ is closer to $mu_{1k}$ than to the other Normal means, then the other components of $X_1$ are also more likely to originate from this same $k$-th element of the mixture.) For instance a most standard way to simulate from (1) is to simulate
          $$Z_1simmathcal M(1;w_1,ldots,w_N)qquad X_1|Z_1=k sim mathcal{N}(x;mu_k, Sigma_k)$$and$$Z_2simmathcal M(1;w_1,ldots,w_N)qquad X_2|Z_2=k sim mathcal{N}(x;mu_k, Sigma_k)$$If the four variables are mutually independent then $X_1$ and $X_2$ are independent.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. What is the meaning of $mathcal{M}$? Can you justify intuitively why the $X_1, X_2$ will most likely to be independent (this is clearer to me), whereas the components of $X_1$ will be dependent (this is not as clear to me)?
            $endgroup$
            – Shamisen Expert
            7 hours ago










          • $begingroup$
            Xi'an is using $mathcal{M}$ to denote the multinomial distribution. With the first parameter $n$ set to 1, the multinomial is also known as the categorical distribution. A categorical random variable takes on one of a finite set of categorical values with probability determined by the second parameter, the $w$ vector. In other words, $Z_i$ represents the group membership of observation $i$.
            $endgroup$
            – mb7744
            59 mins ago


















          5












          $begingroup$

          You are confusing independence between the simulated vectors and independence between the components of a single vector. Unless otherwise stated, two simulations $X_1,X_2$ from$$sumlimits_{i = 1}^N w_i mathcal{N}(x;mu_i, Sigma_i)tag{1}$$will be independent, while the components of the vector $X_1$ will most likely be dependent, even when the matrices $Sigma_i$ are diagonal. (The intuition about that last point is that if $X_{11}$ is closer to $mu_{1k}$ than to the other Normal means, then the other components of $X_1$ are also more likely to originate from this same $k$-th element of the mixture.) For instance a most standard way to simulate from (1) is to simulate
          $$Z_1simmathcal M(1;w_1,ldots,w_N)qquad X_1|Z_1=k sim mathcal{N}(x;mu_k, Sigma_k)$$and$$Z_2simmathcal M(1;w_1,ldots,w_N)qquad X_2|Z_2=k sim mathcal{N}(x;mu_k, Sigma_k)$$If the four variables are mutually independent then $X_1$ and $X_2$ are independent.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your answer. What is the meaning of $mathcal{M}$? Can you justify intuitively why the $X_1, X_2$ will most likely to be independent (this is clearer to me), whereas the components of $X_1$ will be dependent (this is not as clear to me)?
            $endgroup$
            – Shamisen Expert
            7 hours ago










          • $begingroup$
            Xi'an is using $mathcal{M}$ to denote the multinomial distribution. With the first parameter $n$ set to 1, the multinomial is also known as the categorical distribution. A categorical random variable takes on one of a finite set of categorical values with probability determined by the second parameter, the $w$ vector. In other words, $Z_i$ represents the group membership of observation $i$.
            $endgroup$
            – mb7744
            59 mins ago
















          5












          5








          5





          $begingroup$

          You are confusing independence between the simulated vectors and independence between the components of a single vector. Unless otherwise stated, two simulations $X_1,X_2$ from$$sumlimits_{i = 1}^N w_i mathcal{N}(x;mu_i, Sigma_i)tag{1}$$will be independent, while the components of the vector $X_1$ will most likely be dependent, even when the matrices $Sigma_i$ are diagonal. (The intuition about that last point is that if $X_{11}$ is closer to $mu_{1k}$ than to the other Normal means, then the other components of $X_1$ are also more likely to originate from this same $k$-th element of the mixture.) For instance a most standard way to simulate from (1) is to simulate
          $$Z_1simmathcal M(1;w_1,ldots,w_N)qquad X_1|Z_1=k sim mathcal{N}(x;mu_k, Sigma_k)$$and$$Z_2simmathcal M(1;w_1,ldots,w_N)qquad X_2|Z_2=k sim mathcal{N}(x;mu_k, Sigma_k)$$If the four variables are mutually independent then $X_1$ and $X_2$ are independent.






          share|cite|improve this answer











          $endgroup$



          You are confusing independence between the simulated vectors and independence between the components of a single vector. Unless otherwise stated, two simulations $X_1,X_2$ from$$sumlimits_{i = 1}^N w_i mathcal{N}(x;mu_i, Sigma_i)tag{1}$$will be independent, while the components of the vector $X_1$ will most likely be dependent, even when the matrices $Sigma_i$ are diagonal. (The intuition about that last point is that if $X_{11}$ is closer to $mu_{1k}$ than to the other Normal means, then the other components of $X_1$ are also more likely to originate from this same $k$-th element of the mixture.) For instance a most standard way to simulate from (1) is to simulate
          $$Z_1simmathcal M(1;w_1,ldots,w_N)qquad X_1|Z_1=k sim mathcal{N}(x;mu_k, Sigma_k)$$and$$Z_2simmathcal M(1;w_1,ldots,w_N)qquad X_2|Z_2=k sim mathcal{N}(x;mu_k, Sigma_k)$$If the four variables are mutually independent then $X_1$ and $X_2$ are independent.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 6 hours ago

























          answered 7 hours ago









          Xi'anXi'an

          57.9k897360




          57.9k897360












          • $begingroup$
            Thanks for your answer. What is the meaning of $mathcal{M}$? Can you justify intuitively why the $X_1, X_2$ will most likely to be independent (this is clearer to me), whereas the components of $X_1$ will be dependent (this is not as clear to me)?
            $endgroup$
            – Shamisen Expert
            7 hours ago










          • $begingroup$
            Xi'an is using $mathcal{M}$ to denote the multinomial distribution. With the first parameter $n$ set to 1, the multinomial is also known as the categorical distribution. A categorical random variable takes on one of a finite set of categorical values with probability determined by the second parameter, the $w$ vector. In other words, $Z_i$ represents the group membership of observation $i$.
            $endgroup$
            – mb7744
            59 mins ago




















          • $begingroup$
            Thanks for your answer. What is the meaning of $mathcal{M}$? Can you justify intuitively why the $X_1, X_2$ will most likely to be independent (this is clearer to me), whereas the components of $X_1$ will be dependent (this is not as clear to me)?
            $endgroup$
            – Shamisen Expert
            7 hours ago










          • $begingroup$
            Xi'an is using $mathcal{M}$ to denote the multinomial distribution. With the first parameter $n$ set to 1, the multinomial is also known as the categorical distribution. A categorical random variable takes on one of a finite set of categorical values with probability determined by the second parameter, the $w$ vector. In other words, $Z_i$ represents the group membership of observation $i$.
            $endgroup$
            – mb7744
            59 mins ago


















          $begingroup$
          Thanks for your answer. What is the meaning of $mathcal{M}$? Can you justify intuitively why the $X_1, X_2$ will most likely to be independent (this is clearer to me), whereas the components of $X_1$ will be dependent (this is not as clear to me)?
          $endgroup$
          – Shamisen Expert
          7 hours ago




          $begingroup$
          Thanks for your answer. What is the meaning of $mathcal{M}$? Can you justify intuitively why the $X_1, X_2$ will most likely to be independent (this is clearer to me), whereas the components of $X_1$ will be dependent (this is not as clear to me)?
          $endgroup$
          – Shamisen Expert
          7 hours ago












          $begingroup$
          Xi'an is using $mathcal{M}$ to denote the multinomial distribution. With the first parameter $n$ set to 1, the multinomial is also known as the categorical distribution. A categorical random variable takes on one of a finite set of categorical values with probability determined by the second parameter, the $w$ vector. In other words, $Z_i$ represents the group membership of observation $i$.
          $endgroup$
          – mb7744
          59 mins ago






          $begingroup$
          Xi'an is using $mathcal{M}$ to denote the multinomial distribution. With the first parameter $n$ set to 1, the multinomial is also known as the categorical distribution. A categorical random variable takes on one of a finite set of categorical values with probability determined by the second parameter, the $w$ vector. In other words, $Z_i$ represents the group membership of observation $i$.
          $endgroup$
          – mb7744
          59 mins ago




















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