The number of trials which maximizes the expectation of difference between the number of red balls and blue...
$begingroup$
There are $N$ balls with $K$ red balls and $N-K$ blue balls.
It means that the probability that the red ball is drawn with one draw is $K/N$.
There is a random variables
$X$ = the number of red balls drawn.
$Y$ = the number of blue balls drawn.
$Z$ = $X-Y$.
When I draw $M$ balls without replacement from $N$ balls ($K$ red balls and $N-K$ blue balls),
how can I find the proper $M$ which maximizes expectation $mathbb{E}[Z]$?
probability combinatorics
asked Dec 12 '18 at 9:30
kyubkyub
376
$endgroup$
add a comment |
$begingroup$
There are $N$ balls with $K$ red balls and $N-K$ blue balls.
It means that the probability that the red ball is drawn with one draw is $K/N$.
There is a random variables
$X$ = the number of red balls drawn.
$Y$ = the number of blue balls drawn.
$Z$ = $X-Y$.
When I draw $M$ balls without replacement from $N$ balls ($K$ red balls and $N-K$ blue balls),
how can I find the proper $M$ which maximizes expectation $mathbb{E}[Z]$?
probability combinatorics
asked Dec 12 '18 at 9:30
kyubkyub
376
$endgroup$
$begingroup$
What have you tried? Have you tried caculating $Bbb P(Z=i)$ for any $i$ and $M$ for example, or have you tried finding $Bbb E(Z)$ for small values of M?
$endgroup$
– TheD0ubleT
Dec 12 '18 at 9:39
$begingroup$
Since $K<<N$ I assume $K<N/2$ in which case drawing $M>0$ balls without replacement always gives $mathbb{E}[Z]<0$, but $M=0$ gives $mathbb{E}[Z]=0$. Am I missing something?
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:42
$begingroup$
I want to find the value of $M$. If $K<<N$, i think that the proper value of $M$ maximizing $mathbb{E}[Z]$ is $0$. My question is that $M$ which maximizes the $mathbb{E}[Z]$ can be expressed with parameter $N,K$?
$endgroup$
– kyub
Dec 12 '18 at 9:53
1
$begingroup$
It seems the relation $E(Z) = E(X)-E(Y)$ might be relevant here.
$endgroup$
– awkward
Dec 12 '18 at 13:00
add a comment |
$begingroup$
There are $N$ balls with $K$ red balls and $N-K$ blue balls.
It means that the probability that the red ball is drawn with one draw is $K/N$.
There is a random variables
$X$ = the number of red balls drawn.
$Y$ = the number of blue balls drawn.
$Z$ = $X-Y$.
When I draw $M$ balls without replacement from $N$ balls ($K$ red balls and $N-K$ blue balls),
how can I find the proper $M$ which maximizes expectation $mathbb{E}[Z]$?
probability combinatorics
asked Dec 12 '18 at 9:30
kyubkyub
376
$endgroup$
There are $N$ balls with $K$ red balls and $N-K$ blue balls.
It means that the probability that the red ball is drawn with one draw is $K/N$.
There is a random variables
$X$ = the number of red balls drawn.
$Y$ = the number of blue balls drawn.
$Z$ = $X-Y$.
When I draw $M$ balls without replacement from $N$ balls ($K$ red balls and $N-K$ blue balls),
how can I find the proper $M$ which maximizes expectation $mathbb{E}[Z]$?
probability combinatorics
probability combinatorics
asked Dec 12 '18 at 9:30
kyubkyub
376
asked Dec 12 '18 at 9:30
kyubkyub
376
edited Dec 12 '18 at 9:49
kyub
asked Dec 12 '18 at 9:30
kyubkyub
376
asked Dec 12 '18 at 9:30
kyubkyub
376
asked Dec 12 '18 at 9:30
kyubkyub
376
376
$begingroup$
What have you tried? Have you tried caculating $Bbb P(Z=i)$ for any $i$ and $M$ for example, or have you tried finding $Bbb E(Z)$ for small values of M?
$endgroup$
– TheD0ubleT
Dec 12 '18 at 9:39
$begingroup$
Since $K<<N$ I assume $K<N/2$ in which case drawing $M>0$ balls without replacement always gives $mathbb{E}[Z]<0$, but $M=0$ gives $mathbb{E}[Z]=0$. Am I missing something?
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:42
$begingroup$
I want to find the value of $M$. If $K<<N$, i think that the proper value of $M$ maximizing $mathbb{E}[Z]$ is $0$. My question is that $M$ which maximizes the $mathbb{E}[Z]$ can be expressed with parameter $N,K$?
$endgroup$
– kyub
Dec 12 '18 at 9:53
1
$begingroup$
It seems the relation $E(Z) = E(X)-E(Y)$ might be relevant here.
$endgroup$
– awkward
Dec 12 '18 at 13:00
add a comment |
$begingroup$
What have you tried? Have you tried caculating $Bbb P(Z=i)$ for any $i$ and $M$ for example, or have you tried finding $Bbb E(Z)$ for small values of M?
$endgroup$
– TheD0ubleT
Dec 12 '18 at 9:39
$begingroup$
Since $K<<N$ I assume $K<N/2$ in which case drawing $M>0$ balls without replacement always gives $mathbb{E}[Z]<0$, but $M=0$ gives $mathbb{E}[Z]=0$. Am I missing something?
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:42
$begingroup$
I want to find the value of $M$. If $K<<N$, i think that the proper value of $M$ maximizing $mathbb{E}[Z]$ is $0$. My question is that $M$ which maximizes the $mathbb{E}[Z]$ can be expressed with parameter $N,K$?
$endgroup$
– kyub
Dec 12 '18 at 9:53
1
$begingroup$
It seems the relation $E(Z) = E(X)-E(Y)$ might be relevant here.
$endgroup$
– awkward
Dec 12 '18 at 13:00
$begingroup$
What have you tried? Have you tried caculating $Bbb P(Z=i)$ for any $i$ and $M$ for example, or have you tried finding $Bbb E(Z)$ for small values of M?
$endgroup$
– TheD0ubleT
Dec 12 '18 at 9:39
$begingroup$
What have you tried? Have you tried caculating $Bbb P(Z=i)$ for any $i$ and $M$ for example, or have you tried finding $Bbb E(Z)$ for small values of M?
$endgroup$
– TheD0ubleT
Dec 12 '18 at 9:39
$begingroup$
Since $K<<N$ I assume $K<N/2$ in which case drawing $M>0$ balls without replacement always gives $mathbb{E}[Z]<0$, but $M=0$ gives $mathbb{E}[Z]=0$. Am I missing something?
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:42
$begingroup$
Since $K<<N$ I assume $K<N/2$ in which case drawing $M>0$ balls without replacement always gives $mathbb{E}[Z]<0$, but $M=0$ gives $mathbb{E}[Z]=0$. Am I missing something?
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:42
$begingroup$
I want to find the value of $M$. If $K<<N$, i think that the proper value of $M$ maximizing $mathbb{E}[Z]$ is $0$. My question is that $M$ which maximizes the $mathbb{E}[Z]$ can be expressed with parameter $N,K$?
$endgroup$
– kyub
Dec 12 '18 at 9:53
$begingroup$
I want to find the value of $M$. If $K<<N$, i think that the proper value of $M$ maximizing $mathbb{E}[Z]$ is $0$. My question is that $M$ which maximizes the $mathbb{E}[Z]$ can be expressed with parameter $N,K$?
$endgroup$
– kyub
Dec 12 '18 at 9:53
1
1
$begingroup$
It seems the relation $E(Z) = E(X)-E(Y)$ might be relevant here.
$endgroup$
– awkward
Dec 12 '18 at 13:00
$begingroup$
It seems the relation $E(Z) = E(X)-E(Y)$ might be relevant here.
$endgroup$
– awkward
Dec 12 '18 at 13:00
add a comment |
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$begingroup$
What have you tried? Have you tried caculating $Bbb P(Z=i)$ for any $i$ and $M$ for example, or have you tried finding $Bbb E(Z)$ for small values of M?
$endgroup$
– TheD0ubleT
Dec 12 '18 at 9:39
$begingroup$
Since $K<<N$ I assume $K<N/2$ in which case drawing $M>0$ balls without replacement always gives $mathbb{E}[Z]<0$, but $M=0$ gives $mathbb{E}[Z]=0$. Am I missing something?
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:42
$begingroup$
I want to find the value of $M$. If $K<<N$, i think that the proper value of $M$ maximizing $mathbb{E}[Z]$ is $0$. My question is that $M$ which maximizes the $mathbb{E}[Z]$ can be expressed with parameter $N,K$?
$endgroup$
– kyub
Dec 12 '18 at 9:53
1
$begingroup$
It seems the relation $E(Z) = E(X)-E(Y)$ might be relevant here.
$endgroup$
– awkward
Dec 12 '18 at 13:00