Sampling from a continuous population

Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
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 $P(n=i)$?
probability statistics probability-distributions sampling
add a comment |Â
up vote
1
down vote
favorite
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 $P(n=i)$?
probability statistics probability-distributions sampling
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 $P(n=i)$?
probability statistics probability-distributions sampling
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 $P(n=i)$?
probability statistics probability-distributions sampling
edited Aug 17 at 1:14
asked Aug 17 at 0:24
Muriel
297
297
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
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.
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!
â Muriel
Aug 17 at 3:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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!
â Muriel
Aug 17 at 3:16
add a comment |Â
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.
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!
â Muriel
Aug 17 at 3:16
add a comment |Â
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.
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.
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!
â Muriel
Aug 17 at 3:16
add a comment |Â
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!
â 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!
â 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!
â Muriel
Aug 17 at 3:16
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2885287%2fsampling-from-a-continuous-population%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password