Expected number of good presents
$begingroup$
Given $b$ boys and $g$ girls. Children give presents to each other. They know who gives a present whom from random permutation of $1,2,dots b+g$. If child gives present to child with same gender present is good with probability $p_1$, otherwise probability is $p_2$. Presents are good independently. We have a random variable $X$ representing number of good presents. What is $E(X)$ and $Var(X)$?
My idea of solving this:
Without loss of generality I assume that $g le b$. Let $f(k)$ - number of permutations representing cases in which exactly $k$ girls give present to boys. $f(k) = [g (g-1)cdots (g-k+1)b(b-1) cdots (b-k+1)]cdot [g (g-1)cdots (g-k+1)]cdot b! = frac{g!}{(g-k)!}frac{b!}{(b-k)!} frac{g!}{(g-k)!}b!$
$E(X) = sumlimits_{k=0}^{g}{frac{f(k)}{(b+g)!}(E(Binom(b+g-2k, p_1))+E(Binom(2k, p_2)))} = frac{g!g!}{(b+g)!}sumlimits_{k=0}^{g}{frac{(b+g-2k)p_1+(2k)p_2}{(b-k)!(g-k)!(g-k)!}}$
How to finish it?
probability random-variables variance expected-value
asked Dec 11 '18 at 0:00
uuduuudu
12
$endgroup$
add a comment |
$begingroup$
Given $b$ boys and $g$ girls. Children give presents to each other. They know who gives a present whom from random permutation of $1,2,dots b+g$. If child gives present to child with same gender present is good with probability $p_1$, otherwise probability is $p_2$. Presents are good independently. We have a random variable $X$ representing number of good presents. What is $E(X)$ and $Var(X)$?
My idea of solving this:
Without loss of generality I assume that $g le b$. Let $f(k)$ - number of permutations representing cases in which exactly $k$ girls give present to boys. $f(k) = [g (g-1)cdots (g-k+1)b(b-1) cdots (b-k+1)]cdot [g (g-1)cdots (g-k+1)]cdot b! = frac{g!}{(g-k)!}frac{b!}{(b-k)!} frac{g!}{(g-k)!}b!$
$E(X) = sumlimits_{k=0}^{g}{frac{f(k)}{(b+g)!}(E(Binom(b+g-2k, p_1))+E(Binom(2k, p_2)))} = frac{g!g!}{(b+g)!}sumlimits_{k=0}^{g}{frac{(b+g-2k)p_1+(2k)p_2}{(b-k)!(g-k)!(g-k)!}}$
How to finish it?
probability random-variables variance expected-value
asked Dec 11 '18 at 0:00
uuduuudu
12
$endgroup$
add a comment |
$begingroup$
Given $b$ boys and $g$ girls. Children give presents to each other. They know who gives a present whom from random permutation of $1,2,dots b+g$. If child gives present to child with same gender present is good with probability $p_1$, otherwise probability is $p_2$. Presents are good independently. We have a random variable $X$ representing number of good presents. What is $E(X)$ and $Var(X)$?
My idea of solving this:
Without loss of generality I assume that $g le b$. Let $f(k)$ - number of permutations representing cases in which exactly $k$ girls give present to boys. $f(k) = [g (g-1)cdots (g-k+1)b(b-1) cdots (b-k+1)]cdot [g (g-1)cdots (g-k+1)]cdot b! = frac{g!}{(g-k)!}frac{b!}{(b-k)!} frac{g!}{(g-k)!}b!$
$E(X) = sumlimits_{k=0}^{g}{frac{f(k)}{(b+g)!}(E(Binom(b+g-2k, p_1))+E(Binom(2k, p_2)))} = frac{g!g!}{(b+g)!}sumlimits_{k=0}^{g}{frac{(b+g-2k)p_1+(2k)p_2}{(b-k)!(g-k)!(g-k)!}}$
How to finish it?
probability random-variables variance expected-value
asked Dec 11 '18 at 0:00
uuduuudu
12
$endgroup$
Given $b$ boys and $g$ girls. Children give presents to each other. They know who gives a present whom from random permutation of $1,2,dots b+g$. If child gives present to child with same gender present is good with probability $p_1$, otherwise probability is $p_2$. Presents are good independently. We have a random variable $X$ representing number of good presents. What is $E(X)$ and $Var(X)$?
My idea of solving this:
Without loss of generality I assume that $g le b$. Let $f(k)$ - number of permutations representing cases in which exactly $k$ girls give present to boys. $f(k) = [g (g-1)cdots (g-k+1)b(b-1) cdots (b-k+1)]cdot [g (g-1)cdots (g-k+1)]cdot b! = frac{g!}{(g-k)!}frac{b!}{(b-k)!} frac{g!}{(g-k)!}b!$
$E(X) = sumlimits_{k=0}^{g}{frac{f(k)}{(b+g)!}(E(Binom(b+g-2k, p_1))+E(Binom(2k, p_2)))} = frac{g!g!}{(b+g)!}sumlimits_{k=0}^{g}{frac{(b+g-2k)p_1+(2k)p_2}{(b-k)!(g-k)!(g-k)!}}$
How to finish it?
probability random-variables variance expected-value
probability random-variables variance expected-value
asked Dec 11 '18 at 0:00
uuduuudu
12
asked Dec 11 '18 at 0:00
uuduuudu
12
asked Dec 11 '18 at 0:00
uuduuudu
12
asked Dec 11 '18 at 0:00
uuduuudu
12
asked Dec 11 '18 at 0:00
uuduuudu
12
12
add a comment |
add a comment |
1 Answer
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$begingroup$
This is a good example of when using indicator random variables is extremely helpful and avoids the sort of complicated computation that your first attempt involves.
Let $I_i$ be the indicator for the event ${text{the $i$th child received a good present}}$ (that is, $I_i$ equals $1$ if the event holds, and otherwise equals $0$).
Then $X = sum_{i=1}^{b+g} I_i$ so
$$E[X] = Eleft[sum_{i=1}^{b+g} I_i right]= sum_{i=1}^{b+g} E[I_i].$$
It is much easier to compute each $E[I_i]$ individually than to compute $E[X]$ directly.
So, the remaining task is to compute
$$E[I_i] = P(text{the $i$th child received a good present}) = cdots$$
for each $i$. Can you take it from here?
Similarly,
$$E[X^2] = Eleft[left(sum_{i=1}^{b+g} I_iright)^2right] = sum_{i=1}^{b+g} E[I_i^2] + 2 sum_{i ne j} E[I_i I_j].$$
If we compute this, we can compute the variance of $X$.
Note $E[I_i^2] = E[I_i]$ because $I_i$ takes values $0$ or $1$, and note that you have already computed this quantity above.
Thus the only new quantity we have left to compute is $E[I_i I_j]$ for $i ne j$.
This is
$$E[I_i I_j] = Ptext{(both children $i$ and $j$ receive good presents}) = cdots$$
Can you take it from here?
answered Dec 11 '18 at 0:11
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
So, $E(I_i) = frac{g}{g+b}p_1+frac{b}{g+b}p_2$ for $i le g$ and E(I_i) = $frac{g}{g+b}p_2+frac{b}{g+b}p_1$ for $i > g$?
$endgroup$
– uudu
Dec 11 '18 at 0:21
add a comment |
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$begingroup$
This is a good example of when using indicator random variables is extremely helpful and avoids the sort of complicated computation that your first attempt involves.
Let $I_i$ be the indicator for the event ${text{the $i$th child received a good present}}$ (that is, $I_i$ equals $1$ if the event holds, and otherwise equals $0$).
Then $X = sum_{i=1}^{b+g} I_i$ so
$$E[X] = Eleft[sum_{i=1}^{b+g} I_i right]= sum_{i=1}^{b+g} E[I_i].$$
It is much easier to compute each $E[I_i]$ individually than to compute $E[X]$ directly.
So, the remaining task is to compute
$$E[I_i] = P(text{the $i$th child received a good present}) = cdots$$
for each $i$. Can you take it from here?
Similarly,
$$E[X^2] = Eleft[left(sum_{i=1}^{b+g} I_iright)^2right] = sum_{i=1}^{b+g} E[I_i^2] + 2 sum_{i ne j} E[I_i I_j].$$
If we compute this, we can compute the variance of $X$.
Note $E[I_i^2] = E[I_i]$ because $I_i$ takes values $0$ or $1$, and note that you have already computed this quantity above.
Thus the only new quantity we have left to compute is $E[I_i I_j]$ for $i ne j$.
This is
$$E[I_i I_j] = Ptext{(both children $i$ and $j$ receive good presents}) = cdots$$
Can you take it from here?
answered Dec 11 '18 at 0:11
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
So, $E(I_i) = frac{g}{g+b}p_1+frac{b}{g+b}p_2$ for $i le g$ and E(I_i) = $frac{g}{g+b}p_2+frac{b}{g+b}p_1$ for $i > g$?
$endgroup$
– uudu
Dec 11 '18 at 0:21
add a comment |
$begingroup$
This is a good example of when using indicator random variables is extremely helpful and avoids the sort of complicated computation that your first attempt involves.
Let $I_i$ be the indicator for the event ${text{the $i$th child received a good present}}$ (that is, $I_i$ equals $1$ if the event holds, and otherwise equals $0$).
Then $X = sum_{i=1}^{b+g} I_i$ so
$$E[X] = Eleft[sum_{i=1}^{b+g} I_i right]= sum_{i=1}^{b+g} E[I_i].$$
It is much easier to compute each $E[I_i]$ individually than to compute $E[X]$ directly.
So, the remaining task is to compute
$$E[I_i] = P(text{the $i$th child received a good present}) = cdots$$
for each $i$. Can you take it from here?
Similarly,
$$E[X^2] = Eleft[left(sum_{i=1}^{b+g} I_iright)^2right] = sum_{i=1}^{b+g} E[I_i^2] + 2 sum_{i ne j} E[I_i I_j].$$
If we compute this, we can compute the variance of $X$.
Note $E[I_i^2] = E[I_i]$ because $I_i$ takes values $0$ or $1$, and note that you have already computed this quantity above.
Thus the only new quantity we have left to compute is $E[I_i I_j]$ for $i ne j$.
This is
$$E[I_i I_j] = Ptext{(both children $i$ and $j$ receive good presents}) = cdots$$
Can you take it from here?
answered Dec 11 '18 at 0:11
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
So, $E(I_i) = frac{g}{g+b}p_1+frac{b}{g+b}p_2$ for $i le g$ and E(I_i) = $frac{g}{g+b}p_2+frac{b}{g+b}p_1$ for $i > g$?
$endgroup$
– uudu
Dec 11 '18 at 0:21
add a comment |
$begingroup$
This is a good example of when using indicator random variables is extremely helpful and avoids the sort of complicated computation that your first attempt involves.
Let $I_i$ be the indicator for the event ${text{the $i$th child received a good present}}$ (that is, $I_i$ equals $1$ if the event holds, and otherwise equals $0$).
Then $X = sum_{i=1}^{b+g} I_i$ so
$$E[X] = Eleft[sum_{i=1}^{b+g} I_i right]= sum_{i=1}^{b+g} E[I_i].$$
It is much easier to compute each $E[I_i]$ individually than to compute $E[X]$ directly.
So, the remaining task is to compute
$$E[I_i] = P(text{the $i$th child received a good present}) = cdots$$
for each $i$. Can you take it from here?
Similarly,
$$E[X^2] = Eleft[left(sum_{i=1}^{b+g} I_iright)^2right] = sum_{i=1}^{b+g} E[I_i^2] + 2 sum_{i ne j} E[I_i I_j].$$
If we compute this, we can compute the variance of $X$.
Note $E[I_i^2] = E[I_i]$ because $I_i$ takes values $0$ or $1$, and note that you have already computed this quantity above.
Thus the only new quantity we have left to compute is $E[I_i I_j]$ for $i ne j$.
This is
$$E[I_i I_j] = Ptext{(both children $i$ and $j$ receive good presents}) = cdots$$
Can you take it from here?
answered Dec 11 '18 at 0:11
angryavianangryavian
40.6k23380
$endgroup$
This is a good example of when using indicator random variables is extremely helpful and avoids the sort of complicated computation that your first attempt involves.
Let $I_i$ be the indicator for the event ${text{the $i$th child received a good present}}$ (that is, $I_i$ equals $1$ if the event holds, and otherwise equals $0$).
Then $X = sum_{i=1}^{b+g} I_i$ so
$$E[X] = Eleft[sum_{i=1}^{b+g} I_i right]= sum_{i=1}^{b+g} E[I_i].$$
It is much easier to compute each $E[I_i]$ individually than to compute $E[X]$ directly.
So, the remaining task is to compute
$$E[I_i] = P(text{the $i$th child received a good present}) = cdots$$
for each $i$. Can you take it from here?
Similarly,
$$E[X^2] = Eleft[left(sum_{i=1}^{b+g} I_iright)^2right] = sum_{i=1}^{b+g} E[I_i^2] + 2 sum_{i ne j} E[I_i I_j].$$
If we compute this, we can compute the variance of $X$.
Note $E[I_i^2] = E[I_i]$ because $I_i$ takes values $0$ or $1$, and note that you have already computed this quantity above.
Thus the only new quantity we have left to compute is $E[I_i I_j]$ for $i ne j$.
This is
$$E[I_i I_j] = Ptext{(both children $i$ and $j$ receive good presents}) = cdots$$
Can you take it from here?
answered Dec 11 '18 at 0:11
angryavianangryavian
40.6k23380
answered Dec 11 '18 at 0:11
angryavianangryavian
40.6k23380
answered Dec 11 '18 at 0:11
angryavianangryavian
40.6k23380
answered Dec 11 '18 at 0:11
angryavianangryavian
40.6k23380
40.6k23380
$begingroup$
So, $E(I_i) = frac{g}{g+b}p_1+frac{b}{g+b}p_2$ for $i le g$ and E(I_i) = $frac{g}{g+b}p_2+frac{b}{g+b}p_1$ for $i > g$?
$endgroup$
– uudu
Dec 11 '18 at 0:21
add a comment |
$begingroup$
So, $E(I_i) = frac{g}{g+b}p_1+frac{b}{g+b}p_2$ for $i le g$ and E(I_i) = $frac{g}{g+b}p_2+frac{b}{g+b}p_1$ for $i > g$?
$endgroup$
– uudu
Dec 11 '18 at 0:21
$begingroup$
So, $E(I_i) = frac{g}{g+b}p_1+frac{b}{g+b}p_2$ for $i le g$ and E(I_i) = $frac{g}{g+b}p_2+frac{b}{g+b}p_1$ for $i > g$?
$endgroup$
– uudu
Dec 11 '18 at 0:21
$begingroup$
So, $E(I_i) = frac{g}{g+b}p_1+frac{b}{g+b}p_2$ for $i le g$ and E(I_i) = $frac{g}{g+b}p_2+frac{b}{g+b}p_1$ for $i > g$?
$endgroup$
– uudu
Dec 11 '18 at 0:21
add a comment |
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