Sampling from a continuous population

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Suppose I have a volume $V$ (ml) of a solution, containing a fixed number $n$ of particles of a certain substance (included in the volume count). Assume also we don't know anything about the particles spatial distribution in the volume.



If I take a random sample of volume $V_sample$ from my solution, and call $Y$ the number of substance particles in the sample, how can I compute the probability distribution of $Y$?



And what if the original number $n$ is not deterministic but itself distributed with some known pmf $P(n=i)$?







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    Suppose I have a volume $V$ (ml) of a solution, containing a fixed number $n$ of particles of a certain substance (included in the volume count). Assume also we don't know anything about the particles spatial distribution in the volume.



    If I take a random sample of volume $V_sample$ from my solution, and call $Y$ the number of substance particles in the sample, how can I compute the probability distribution of $Y$?



    And what if the original number $n$ is not deterministic but itself distributed with some known pmf $P(n=i)$?







    share|cite|improve this question
























      up vote
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      down vote

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      1









      up vote
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      Suppose I have a volume $V$ (ml) of a solution, containing a fixed number $n$ of particles of a certain substance (included in the volume count). Assume also we don't know anything about the particles spatial distribution in the volume.



      If I take a random sample of volume $V_sample$ from my solution, and call $Y$ the number of substance particles in the sample, how can I compute the probability distribution of $Y$?



      And what if the original number $n$ is not deterministic but itself distributed with some known pmf $P(n=i)$?







      share|cite|improve this question














      Suppose I have a volume $V$ (ml) of a solution, containing a fixed number $n$ of particles of a certain substance (included in the volume count). Assume also we don't know anything about the particles spatial distribution in the volume.



      If I take a random sample of volume $V_sample$ from my solution, and call $Y$ the number of substance particles in the sample, how can I compute the probability distribution of $Y$?



      And what if the original number $n$ is not deterministic but itself distributed with some known pmf $P(n=i)$?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 17 at 1:14

























      asked Aug 17 at 0:24









      Muriel

      297




      297




















          1 Answer
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          Since you're asking this on the math site and not on the physics site, I'm assuming that you're abstracting from physical effects like excluded volume and are assuming an ideal solution in which each solvent particle is independently uniformly distributed over the entire accessible volume.



          Each solvent particle is in the sample volume $V_textsample$ with independent probability $p=V_textsample/V$. Then the number $Y$ of solvent particles in the sample volume $V_textsample$ is binomially distributed with parameters $n$ and $p$:



          $$
          mathsf P(Y=y)=binom nyp^y(1-p)^n-y=binom nyleft(fracV_textsampleVright)^yleft(1-fracV_textsampleVright)^n-y;.$$



          For the large numbers of particles contained in typical volumes, this binomial distribution is very well approximated by a normal distribution with the same mean and variance. The mean is $mu=np=nfracV_textsampleV$, and the variance is $sigma^2=np(1-p)=nfracV_textsampleVleft(1-fracV_textsampleVright)$.



          If the number of particles is itself a random variable $N$, you need to apply the law of total probability:



          $$
          mathsf P(Y=y)=sum_n=y^inftymathsf P(N=n)mathsf P(Y=ymid N=n);,
          $$



          where the conditional probability $mathsf P(Y=ymid N=n)$ is given by the expression for $mathsf P(Y=y)$ in the first part of the answer.






          share|cite|improve this answer




















          • Definitely easier than I thought, I got lost into the details of my problem. And yeah I was just looking for an abstract solution to it: thank you very much!
            – Muriel
            Aug 17 at 3:16










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          1 Answer
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          1 Answer
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          active

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          up vote
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          accepted










          Since you're asking this on the math site and not on the physics site, I'm assuming that you're abstracting from physical effects like excluded volume and are assuming an ideal solution in which each solvent particle is independently uniformly distributed over the entire accessible volume.



          Each solvent particle is in the sample volume $V_textsample$ with independent probability $p=V_textsample/V$. Then the number $Y$ of solvent particles in the sample volume $V_textsample$ is binomially distributed with parameters $n$ and $p$:



          $$
          mathsf P(Y=y)=binom nyp^y(1-p)^n-y=binom nyleft(fracV_textsampleVright)^yleft(1-fracV_textsampleVright)^n-y;.$$



          For the large numbers of particles contained in typical volumes, this binomial distribution is very well approximated by a normal distribution with the same mean and variance. The mean is $mu=np=nfracV_textsampleV$, and the variance is $sigma^2=np(1-p)=nfracV_textsampleVleft(1-fracV_textsampleVright)$.



          If the number of particles is itself a random variable $N$, you need to apply the law of total probability:



          $$
          mathsf P(Y=y)=sum_n=y^inftymathsf P(N=n)mathsf P(Y=ymid N=n);,
          $$



          where the conditional probability $mathsf P(Y=ymid N=n)$ is given by the expression for $mathsf P(Y=y)$ in the first part of the answer.






          share|cite|improve this answer




















          • Definitely easier than I thought, I got lost into the details of my problem. And yeah I was just looking for an abstract solution to it: thank you very much!
            – Muriel
            Aug 17 at 3:16














          up vote
          1
          down vote



          accepted










          Since you're asking this on the math site and not on the physics site, I'm assuming that you're abstracting from physical effects like excluded volume and are assuming an ideal solution in which each solvent particle is independently uniformly distributed over the entire accessible volume.



          Each solvent particle is in the sample volume $V_textsample$ with independent probability $p=V_textsample/V$. Then the number $Y$ of solvent particles in the sample volume $V_textsample$ is binomially distributed with parameters $n$ and $p$:



          $$
          mathsf P(Y=y)=binom nyp^y(1-p)^n-y=binom nyleft(fracV_textsampleVright)^yleft(1-fracV_textsampleVright)^n-y;.$$



          For the large numbers of particles contained in typical volumes, this binomial distribution is very well approximated by a normal distribution with the same mean and variance. The mean is $mu=np=nfracV_textsampleV$, and the variance is $sigma^2=np(1-p)=nfracV_textsampleVleft(1-fracV_textsampleVright)$.



          If the number of particles is itself a random variable $N$, you need to apply the law of total probability:



          $$
          mathsf P(Y=y)=sum_n=y^inftymathsf P(N=n)mathsf P(Y=ymid N=n);,
          $$



          where the conditional probability $mathsf P(Y=ymid N=n)$ is given by the expression for $mathsf P(Y=y)$ in the first part of the answer.






          share|cite|improve this answer




















          • Definitely easier than I thought, I got lost into the details of my problem. And yeah I was just looking for an abstract solution to it: thank you very much!
            – Muriel
            Aug 17 at 3:16












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Since you're asking this on the math site and not on the physics site, I'm assuming that you're abstracting from physical effects like excluded volume and are assuming an ideal solution in which each solvent particle is independently uniformly distributed over the entire accessible volume.



          Each solvent particle is in the sample volume $V_textsample$ with independent probability $p=V_textsample/V$. Then the number $Y$ of solvent particles in the sample volume $V_textsample$ is binomially distributed with parameters $n$ and $p$:



          $$
          mathsf P(Y=y)=binom nyp^y(1-p)^n-y=binom nyleft(fracV_textsampleVright)^yleft(1-fracV_textsampleVright)^n-y;.$$



          For the large numbers of particles contained in typical volumes, this binomial distribution is very well approximated by a normal distribution with the same mean and variance. The mean is $mu=np=nfracV_textsampleV$, and the variance is $sigma^2=np(1-p)=nfracV_textsampleVleft(1-fracV_textsampleVright)$.



          If the number of particles is itself a random variable $N$, you need to apply the law of total probability:



          $$
          mathsf P(Y=y)=sum_n=y^inftymathsf P(N=n)mathsf P(Y=ymid N=n);,
          $$



          where the conditional probability $mathsf P(Y=ymid N=n)$ is given by the expression for $mathsf P(Y=y)$ in the first part of the answer.






          share|cite|improve this answer












          Since you're asking this on the math site and not on the physics site, I'm assuming that you're abstracting from physical effects like excluded volume and are assuming an ideal solution in which each solvent particle is independently uniformly distributed over the entire accessible volume.



          Each solvent particle is in the sample volume $V_textsample$ with independent probability $p=V_textsample/V$. Then the number $Y$ of solvent particles in the sample volume $V_textsample$ is binomially distributed with parameters $n$ and $p$:



          $$
          mathsf P(Y=y)=binom nyp^y(1-p)^n-y=binom nyleft(fracV_textsampleVright)^yleft(1-fracV_textsampleVright)^n-y;.$$



          For the large numbers of particles contained in typical volumes, this binomial distribution is very well approximated by a normal distribution with the same mean and variance. The mean is $mu=np=nfracV_textsampleV$, and the variance is $sigma^2=np(1-p)=nfracV_textsampleVleft(1-fracV_textsampleVright)$.



          If the number of particles is itself a random variable $N$, you need to apply the law of total probability:



          $$
          mathsf P(Y=y)=sum_n=y^inftymathsf P(N=n)mathsf P(Y=ymid N=n);,
          $$



          where the conditional probability $mathsf P(Y=ymid N=n)$ is given by the expression for $mathsf P(Y=y)$ in the first part of the answer.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 17 at 1:24









          joriki

          165k10180330




          165k10180330











          • Definitely easier than I thought, I got lost into the details of my problem. And yeah I was just looking for an abstract solution to it: thank you very much!
            – Muriel
            Aug 17 at 3:16
















          • Definitely easier than I thought, I got lost into the details of my problem. And yeah I was just looking for an abstract solution to it: thank you very much!
            – Muriel
            Aug 17 at 3:16















          Definitely easier than I thought, I got lost into the details of my problem. And yeah I was just looking for an abstract solution to it: thank you very much!
          – Muriel
          Aug 17 at 3:16




          Definitely easier than I thought, I got lost into the details of my problem. And yeah I was just looking for an abstract solution to it: thank you very much!
          – Muriel
          Aug 17 at 3:16












           

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