Mixing
![let $X$ be an arbitrary Hausdroff space, then is $Xsetminus{x_0}$ always dense? [closed]](https://lh4.googleusercontent.com/-N2Jq52TCMZ4/AAAAAAAAAAI/AAAAAAAAAAk/78-FJBqBAwg/s72-c/photo.jpg?sz=32)
Given $Xsimtext{Binomial}(m,p)$ and $Ysimtext{Binomial}(n,p)$, calculate $textbf{P}(X = x | X + Y = s)$
$begingroup$
Let $X$ and $Y$ be independent RV's such that $Xsimtext{Binomial}(m,p)$ and $Ysimtext{Binomial}(n,p)$. Determine
(a) $textbf{P}(X = x mid X+Y = s)$
(b) $textbf{E}(Xmid X + Y)$ and $textbf{V}(X mid X+Y)$
(c) Check that $textbf{E}{textbf{E}(Xmid X+Y)} = textbf{E}(X)$
MY ATTEMPT
Firsly, I would like to determine the distribution of $Z = X+Y$. Since $X$ and $Y$ are independent,
begin{align*}
textbf{P}(Z = z) = textbf{P}(X + Y = z) = sum_{x=0}^{z}textbf{P}(X = x,Y = z - x) = sum_{x=0}^{z}textbf{P}(X = x)textbf{P}(Y = z-x)
end{align*}
Consequently, we have
begin{align*}
p_{Z}(z) & = textbf{P}(Z = z) = sum_{x=0}^{z}{nchoose z-x}p^{z-x}(1-p)^{n-z+x}{mchoose x}p^{x}(1-p)^{m-x}\
& = sum_{x=0}^{z}{nchoose z-x}{mchoose x}p^{z}(1-p)^{m + n - z} = {m+nchoose z}p^{z}(1-p)^{m+n-z}
end{align*}
Another possible approach is to consider each Binomial distribution as a sum of Bernoulli distributions. Therefore $Zsimtext{Binomial}(m+n,p)$. Based on this, could someone help me out?
EDIT
(a) As a consequence from the independence between $X$ and $Y$, we have
begin{align*}
textbf{P}(X = x | X + Y = s) & = frac{textbf{P}(X = x, X+Y = s)}{textbf{P}(X+Y = s)} = frac{textbf{P}(X = x)textbf{P}(Y = s - x)}{textbf{P}(X+Y = s)}\\
& = frac{displaystyle{mchoose x}{nchoose s-x}p^{s}(1-p)^{m+n-s}}{displaystyle{m+nchoose s}p^{s}(1-p)^{m+n-s}} = frac{displaystyle{mchoose x}{nchoose s - x}}{displaystyle{m+nchoose s}}
end{align*}
(b) According to wikipedia (Hypergeometric random variable), we have
begin{align*}
textbf{E}(X | X + Y) = frac{sm}{m+n}quadtext{and}quadtextbf{V}(X | X+Y) = ldots
end{align*}
(c) Finally, we have
begin{align*}
textbf{E}{textbf{E}(X|X+Y)} = frac{m}{m+n}textbf{E}(X+Y) = frac{m(m+n)p}{m+n} = mp = textbf{E}(X)
end{align*}
probability probability-theory conditional-expectation variance expected-value
asked Jan 26 at 20:30
user1337user1337
46310
$endgroup$
|
show 1 more comment
$begingroup$
Let $X$ and $Y$ be independent RV's such that $Xsimtext{Binomial}(m,p)$ and $Ysimtext{Binomial}(n,p)$. Determine
(a) $textbf{P}(X = x mid X+Y = s)$
(b) $textbf{E}(Xmid X + Y)$ and $textbf{V}(X mid X+Y)$
(c) Check that $textbf{E}{textbf{E}(Xmid X+Y)} = textbf{E}(X)$
MY ATTEMPT
Firsly, I would like to determine the distribution of $Z = X+Y$. Since $X$ and $Y$ are independent,
begin{align*}
textbf{P}(Z = z) = textbf{P}(X + Y = z) = sum_{x=0}^{z}textbf{P}(X = x,Y = z - x) = sum_{x=0}^{z}textbf{P}(X = x)textbf{P}(Y = z-x)
end{align*}
Consequently, we have
begin{align*}
p_{Z}(z) & = textbf{P}(Z = z) = sum_{x=0}^{z}{nchoose z-x}p^{z-x}(1-p)^{n-z+x}{mchoose x}p^{x}(1-p)^{m-x}\
& = sum_{x=0}^{z}{nchoose z-x}{mchoose x}p^{z}(1-p)^{m + n - z} = {m+nchoose z}p^{z}(1-p)^{m+n-z}
end{align*}
Another possible approach is to consider each Binomial distribution as a sum of Bernoulli distributions. Therefore $Zsimtext{Binomial}(m+n,p)$. Based on this, could someone help me out?
EDIT
(a) As a consequence from the independence between $X$ and $Y$, we have
begin{align*}
textbf{P}(X = x | X + Y = s) & = frac{textbf{P}(X = x, X+Y = s)}{textbf{P}(X+Y = s)} = frac{textbf{P}(X = x)textbf{P}(Y = s - x)}{textbf{P}(X+Y = s)}\\
& = frac{displaystyle{mchoose x}{nchoose s-x}p^{s}(1-p)^{m+n-s}}{displaystyle{m+nchoose s}p^{s}(1-p)^{m+n-s}} = frac{displaystyle{mchoose x}{nchoose s - x}}{displaystyle{m+nchoose s}}
end{align*}
(b) According to wikipedia (Hypergeometric random variable), we have
begin{align*}
textbf{E}(X | X + Y) = frac{sm}{m+n}quadtext{and}quadtextbf{V}(X | X+Y) = ldots
end{align*}
(c) Finally, we have
begin{align*}
textbf{E}{textbf{E}(X|X+Y)} = frac{m}{m+n}textbf{E}(X+Y) = frac{m(m+n)p}{m+n} = mp = textbf{E}(X)
end{align*}
probability probability-theory conditional-expectation variance expected-value
asked Jan 26 at 20:30
user1337user1337
46310
$endgroup$
2
$begingroup$
You definitely can consider each binomial as the sum of bernoulli's. If you choose to go the way of densities, what will really help is to consider Vandermonde’s identity (first discovered by Zhu Shijie in 1303 in Ancient China): $$ sum_{j=0}^k{nchoose k-j} {mchoose j}= {m+n choose k} $$
$endgroup$
– user321627
Jan 26 at 23:38
$begingroup$
I would also check the summation indices you have above.
$endgroup$
– user321627
Jan 27 at 0:15
$begingroup$
There are some conceptual mistakes in your attempt and edit. The second equality in the "My Attempt" is wrong (what you get next would be correct, using Vandermonde's identity, but that's not a consequence of what you wrote just before). The same is true also for the first equality in "My Edit" (here, use first the definition of conditional probability).
$endgroup$
– user52227
Jan 27 at 13:33
$begingroup$
Thanks for the contributions, user52227. I think I have implemented them correctly this time. By the way, could you provide an answer to the last updated question?
$endgroup$
– user1337
Jan 27 at 18:32
$begingroup$
Where's the last updated question?
$endgroup$
– user321627
Jan 27 at 21:41
|
show 1 more comment
$begingroup$
Let $X$ and $Y$ be independent RV's such that $Xsimtext{Binomial}(m,p)$ and $Ysimtext{Binomial}(n,p)$. Determine
(a) $textbf{P}(X = x mid X+Y = s)$
(b) $textbf{E}(Xmid X + Y)$ and $textbf{V}(X mid X+Y)$
(c) Check that $textbf{E}{textbf{E}(Xmid X+Y)} = textbf{E}(X)$
MY ATTEMPT
Firsly, I would like to determine the distribution of $Z = X+Y$. Since $X$ and $Y$ are independent,
begin{align*}
textbf{P}(Z = z) = textbf{P}(X + Y = z) = sum_{x=0}^{z}textbf{P}(X = x,Y = z - x) = sum_{x=0}^{z}textbf{P}(X = x)textbf{P}(Y = z-x)
end{align*}
Consequently, we have
begin{align*}
p_{Z}(z) & = textbf{P}(Z = z) = sum_{x=0}^{z}{nchoose z-x}p^{z-x}(1-p)^{n-z+x}{mchoose x}p^{x}(1-p)^{m-x}\
& = sum_{x=0}^{z}{nchoose z-x}{mchoose x}p^{z}(1-p)^{m + n - z} = {m+nchoose z}p^{z}(1-p)^{m+n-z}
end{align*}
Another possible approach is to consider each Binomial distribution as a sum of Bernoulli distributions. Therefore $Zsimtext{Binomial}(m+n,p)$. Based on this, could someone help me out?
EDIT
(a) As a consequence from the independence between $X$ and $Y$, we have
begin{align*}
textbf{P}(X = x | X + Y = s) & = frac{textbf{P}(X = x, X+Y = s)}{textbf{P}(X+Y = s)} = frac{textbf{P}(X = x)textbf{P}(Y = s - x)}{textbf{P}(X+Y = s)}\\
& = frac{displaystyle{mchoose x}{nchoose s-x}p^{s}(1-p)^{m+n-s}}{displaystyle{m+nchoose s}p^{s}(1-p)^{m+n-s}} = frac{displaystyle{mchoose x}{nchoose s - x}}{displaystyle{m+nchoose s}}
end{align*}
(b) According to wikipedia (Hypergeometric random variable), we have
begin{align*}
textbf{E}(X | X + Y) = frac{sm}{m+n}quadtext{and}quadtextbf{V}(X | X+Y) = ldots
end{align*}
(c) Finally, we have
begin{align*}
textbf{E}{textbf{E}(X|X+Y)} = frac{m}{m+n}textbf{E}(X+Y) = frac{m(m+n)p}{m+n} = mp = textbf{E}(X)
end{align*}
probability probability-theory conditional-expectation variance expected-value
asked Jan 26 at 20:30
user1337user1337
46310
$endgroup$
Let $X$ and $Y$ be independent RV's such that $Xsimtext{Binomial}(m,p)$ and $Ysimtext{Binomial}(n,p)$. Determine
(a) $textbf{P}(X = x mid X+Y = s)$
(b) $textbf{E}(Xmid X + Y)$ and $textbf{V}(X mid X+Y)$
(c) Check that $textbf{E}{textbf{E}(Xmid X+Y)} = textbf{E}(X)$
MY ATTEMPT
Firsly, I would like to determine the distribution of $Z = X+Y$. Since $X$ and $Y$ are independent,
begin{align*}
textbf{P}(Z = z) = textbf{P}(X + Y = z) = sum_{x=0}^{z}textbf{P}(X = x,Y = z - x) = sum_{x=0}^{z}textbf{P}(X = x)textbf{P}(Y = z-x)
end{align*}
Consequently, we have
begin{align*}
p_{Z}(z) & = textbf{P}(Z = z) = sum_{x=0}^{z}{nchoose z-x}p^{z-x}(1-p)^{n-z+x}{mchoose x}p^{x}(1-p)^{m-x}\
& = sum_{x=0}^{z}{nchoose z-x}{mchoose x}p^{z}(1-p)^{m + n - z} = {m+nchoose z}p^{z}(1-p)^{m+n-z}
end{align*}
Another possible approach is to consider each Binomial distribution as a sum of Bernoulli distributions. Therefore $Zsimtext{Binomial}(m+n,p)$. Based on this, could someone help me out?
EDIT
(a) As a consequence from the independence between $X$ and $Y$, we have
begin{align*}
textbf{P}(X = x | X + Y = s) & = frac{textbf{P}(X = x, X+Y = s)}{textbf{P}(X+Y = s)} = frac{textbf{P}(X = x)textbf{P}(Y = s - x)}{textbf{P}(X+Y = s)}\\
& = frac{displaystyle{mchoose x}{nchoose s-x}p^{s}(1-p)^{m+n-s}}{displaystyle{m+nchoose s}p^{s}(1-p)^{m+n-s}} = frac{displaystyle{mchoose x}{nchoose s - x}}{displaystyle{m+nchoose s}}
end{align*}
(b) According to wikipedia (Hypergeometric random variable), we have
begin{align*}
textbf{E}(X | X + Y) = frac{sm}{m+n}quadtext{and}quadtextbf{V}(X | X+Y) = ldots
end{align*}
(c) Finally, we have
begin{align*}
textbf{E}{textbf{E}(X|X+Y)} = frac{m}{m+n}textbf{E}(X+Y) = frac{m(m+n)p}{m+n} = mp = textbf{E}(X)
end{align*}
probability probability-theory conditional-expectation variance expected-value
probability probability-theory conditional-expectation variance expected-value
asked Jan 26 at 20:30
user1337user1337
46310
asked Jan 26 at 20:30
user1337user1337
46310
edited Jan 27 at 20:40
user1337
asked Jan 26 at 20:30
user1337user1337
46310
asked Jan 26 at 20:30
user1337user1337
46310
asked Jan 26 at 20:30
user1337user1337
46310
46310
2
$begingroup$
You definitely can consider each binomial as the sum of bernoulli's. If you choose to go the way of densities, what will really help is to consider Vandermonde’s identity (first discovered by Zhu Shijie in 1303 in Ancient China): $$ sum_{j=0}^k{nchoose k-j} {mchoose j}= {m+n choose k} $$
$endgroup$
– user321627
Jan 26 at 23:38
$begingroup$
I would also check the summation indices you have above.
$endgroup$
– user321627
Jan 27 at 0:15
$begingroup$
There are some conceptual mistakes in your attempt and edit. The second equality in the "My Attempt" is wrong (what you get next would be correct, using Vandermonde's identity, but that's not a consequence of what you wrote just before). The same is true also for the first equality in "My Edit" (here, use first the definition of conditional probability).
$endgroup$
– user52227
Jan 27 at 13:33
$begingroup$
Thanks for the contributions, user52227. I think I have implemented them correctly this time. By the way, could you provide an answer to the last updated question?
$endgroup$
– user1337
Jan 27 at 18:32
$begingroup$
Where's the last updated question?
$endgroup$
– user321627
Jan 27 at 21:41
|
show 1 more comment
2
$begingroup$
You definitely can consider each binomial as the sum of bernoulli's. If you choose to go the way of densities, what will really help is to consider Vandermonde’s identity (first discovered by Zhu Shijie in 1303 in Ancient China): $$ sum_{j=0}^k{nchoose k-j} {mchoose j}= {m+n choose k} $$
$endgroup$
– user321627
Jan 26 at 23:38
$begingroup$
I would also check the summation indices you have above.
$endgroup$
– user321627
Jan 27 at 0:15
$begingroup$
There are some conceptual mistakes in your attempt and edit. The second equality in the "My Attempt" is wrong (what you get next would be correct, using Vandermonde's identity, but that's not a consequence of what you wrote just before). The same is true also for the first equality in "My Edit" (here, use first the definition of conditional probability).
$endgroup$
– user52227
Jan 27 at 13:33
$begingroup$
Thanks for the contributions, user52227. I think I have implemented them correctly this time. By the way, could you provide an answer to the last updated question?
$endgroup$
– user1337
Jan 27 at 18:32
$begingroup$
Where's the last updated question?
$endgroup$
– user321627
Jan 27 at 21:41
2
2
$begingroup$
You definitely can consider each binomial as the sum of bernoulli's. If you choose to go the way of densities, what will really help is to consider Vandermonde’s identity (first discovered by Zhu Shijie in 1303 in Ancient China): $$ sum_{j=0}^k{nchoose k-j} {mchoose j}= {m+n choose k} $$
$endgroup$
– user321627
Jan 26 at 23:38
$begingroup$
You definitely can consider each binomial as the sum of bernoulli's. If you choose to go the way of densities, what will really help is to consider Vandermonde’s identity (first discovered by Zhu Shijie in 1303 in Ancient China): $$ sum_{j=0}^k{nchoose k-j} {mchoose j}= {m+n choose k} $$
$endgroup$
– user321627
Jan 26 at 23:38
$begingroup$
I would also check the summation indices you have above.
$endgroup$
– user321627
Jan 27 at 0:15
$begingroup$
I would also check the summation indices you have above.
$endgroup$
– user321627
Jan 27 at 0:15
$begingroup$
There are some conceptual mistakes in your attempt and edit. The second equality in the "My Attempt" is wrong (what you get next would be correct, using Vandermonde's identity, but that's not a consequence of what you wrote just before). The same is true also for the first equality in "My Edit" (here, use first the definition of conditional probability).
$endgroup$
– user52227
Jan 27 at 13:33
$begingroup$
There are some conceptual mistakes in your attempt and edit. The second equality in the "My Attempt" is wrong (what you get next would be correct, using Vandermonde's identity, but that's not a consequence of what you wrote just before). The same is true also for the first equality in "My Edit" (here, use first the definition of conditional probability).
$endgroup$
– user52227
Jan 27 at 13:33
$begingroup$
Thanks for the contributions, user52227. I think I have implemented them correctly this time. By the way, could you provide an answer to the last updated question?
$endgroup$
– user1337
Jan 27 at 18:32
$begingroup$
Thanks for the contributions, user52227. I think I have implemented them correctly this time. By the way, could you provide an answer to the last updated question?
$endgroup$
– user1337
Jan 27 at 18:32
$begingroup$
Where's the last updated question?
$endgroup$
– user321627
Jan 27 at 21:41
$begingroup$
Where's the last updated question?
$endgroup$
– user321627
Jan 27 at 21:41
|
show 1 more comment
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$begingroup$
You definitely can consider each binomial as the sum of bernoulli's. If you choose to go the way of densities, what will really help is to consider Vandermonde’s identity (first discovered by Zhu Shijie in 1303 in Ancient China): $$ sum_{j=0}^k{nchoose k-j} {mchoose j}= {m+n choose k} $$
$endgroup$
– user321627
Jan 26 at 23:38
$begingroup$
I would also check the summation indices you have above.
$endgroup$
– user321627
Jan 27 at 0:15
$begingroup$
There are some conceptual mistakes in your attempt and edit. The second equality in the "My Attempt" is wrong (what you get next would be correct, using Vandermonde's identity, but that's not a consequence of what you wrote just before). The same is true also for the first equality in "My Edit" (here, use first the definition of conditional probability).
$endgroup$
– user52227
Jan 27 at 13:33
$begingroup$
Thanks for the contributions, user52227. I think I have implemented them correctly this time. By the way, could you provide an answer to the last updated question?
$endgroup$
– user1337
Jan 27 at 18:32
$begingroup$
Where's the last updated question?
$endgroup$
– user321627
Jan 27 at 21:41