Expected Value for defined Random Variables $X$ and $Y$
$begingroup$
I don't understand what the random variables here mean exactly. $B$ is a $5$ element subset of the set $A$.
$X$ over here is defined by $B$ intersection ${a,b,c,d}$. Does this mean that ${a,b,c,d}$ is part of the $5$ element subset of $B$? I'm assuming I have to use the indicator variables to solve this.
If I define $X = 1$ as $B$ intersection ${a,b,c,d}$,
so $E(X) = P(X=1) = $${5}choose{2}$ * ${4}choose{2}$ / ${8}choose{2}$ ?
I know for independence, $P(X=x)*P(Y=y) = P(X=x $ $ and $ $ Y=y)$
But again, I'm not sure what the defined random variables mean, how should I solve this problem, it's been a struggle for the last couple of hours.
probability probability-theory discrete-mathematics random-variables expected-value
edited Dec 17 '18 at 5:51
dmtri
1,5082521
asked Dec 17 '18 at 5:10
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
I don't understand what the random variables here mean exactly. $B$ is a $5$ element subset of the set $A$.
$X$ over here is defined by $B$ intersection ${a,b,c,d}$. Does this mean that ${a,b,c,d}$ is part of the $5$ element subset of $B$? I'm assuming I have to use the indicator variables to solve this.
If I define $X = 1$ as $B$ intersection ${a,b,c,d}$,
so $E(X) = P(X=1) = $${5}choose{2}$ * ${4}choose{2}$ / ${8}choose{2}$ ?
I know for independence, $P(X=x)*P(Y=y) = P(X=x $ $ and $ $ Y=y)$
But again, I'm not sure what the defined random variables mean, how should I solve this problem, it's been a struggle for the last couple of hours.
probability probability-theory discrete-mathematics random-variables expected-value
edited Dec 17 '18 at 5:51
dmtri
1,5082521
asked Dec 17 '18 at 5:10
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
I don't understand what the random variables here mean exactly. $B$ is a $5$ element subset of the set $A$.
$X$ over here is defined by $B$ intersection ${a,b,c,d}$. Does this mean that ${a,b,c,d}$ is part of the $5$ element subset of $B$? I'm assuming I have to use the indicator variables to solve this.
If I define $X = 1$ as $B$ intersection ${a,b,c,d}$,
so $E(X) = P(X=1) = $${5}choose{2}$ * ${4}choose{2}$ / ${8}choose{2}$ ?
I know for independence, $P(X=x)*P(Y=y) = P(X=x $ $ and $ $ Y=y)$
But again, I'm not sure what the defined random variables mean, how should I solve this problem, it's been a struggle for the last couple of hours.
probability probability-theory discrete-mathematics random-variables expected-value
edited Dec 17 '18 at 5:51
dmtri
1,5082521
asked Dec 17 '18 at 5:10
TobyToby
1577
$endgroup$
I don't understand what the random variables here mean exactly. $B$ is a $5$ element subset of the set $A$.
$X$ over here is defined by $B$ intersection ${a,b,c,d}$. Does this mean that ${a,b,c,d}$ is part of the $5$ element subset of $B$? I'm assuming I have to use the indicator variables to solve this.
If I define $X = 1$ as $B$ intersection ${a,b,c,d}$,
so $E(X) = P(X=1) = $${5}choose{2}$ * ${4}choose{2}$ / ${8}choose{2}$ ?
I know for independence, $P(X=x)*P(Y=y) = P(X=x $ $ and $ $ Y=y)$
But again, I'm not sure what the defined random variables mean, how should I solve this problem, it's been a struggle for the last couple of hours.
probability probability-theory discrete-mathematics random-variables expected-value
probability probability-theory discrete-mathematics random-variables expected-value
edited Dec 17 '18 at 5:51
dmtri
1,5082521
asked Dec 17 '18 at 5:10
TobyToby
1577
edited Dec 17 '18 at 5:51
dmtri
1,5082521
asked Dec 17 '18 at 5:10
TobyToby
1577
edited Dec 17 '18 at 5:51
dmtri
1,5082521
edited Dec 17 '18 at 5:51
dmtri
1,5082521
edited Dec 17 '18 at 5:51
dmtri
1,5082521
1,5082521
asked Dec 17 '18 at 5:10
TobyToby
1577
asked Dec 17 '18 at 5:10
TobyToby
1577
asked Dec 17 '18 at 5:10
TobyToby
1577
1577
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As is pointed out, $|A|$ denotes the cardinality (the number of elements) of $A$. Thus we may write
$$
X = sum_{x=a,b,c,d} 1_{x in B},$$ and $$Y = sum_{x=e,f,g,h}1_{xin B}$$ where $1_{zin E}$ denotes the indicator function (taking values in ${0,1}$ whose value is $1$ if and only if the $zin E$.) We can immediately observe that
$$
X+Y =|B| = 5.
$$ This readily implies that $X$ and $Y$ are dependent. And since expectation operation is linear, we have
$$
E[X] = sum_{x=a,b,c,d} E[1_{x in B}] = sum_{x=a,b,c,d} P(x in B).
$$ For arbitrary $x$, it holds that
$$
P(xin B) = frac{binom{7}{4}}{binom{5}{8}} = frac{5}{8}.
$$Hence, this gives
$$
E[X] = 4cdotfrac{5}{8} = frac{5}{2},
$$ and $$
E[Y] =frac{5}{2}$$ also follows.
answered Dec 17 '18 at 5:34
SongSong
14.6k1635
$endgroup$
$begingroup$
I'm a little lost on the 1 z is an element in E part. What does that mean the indicator function taking values 0,1? Also, how is X+Y=|B|? Shouldn't it be equal to A because A has {a.b.c.d.e.f.g}? And where is 5Choose8 coming from?
$endgroup$
– Toby
Dec 17 '18 at 5:46
1
$begingroup$
I used $z$ and $E$ as generic variables. The Indicator function $1_{zin E}$, also called characteristic function, is a $0$-$1$ function which has value $1$ if $zin E$ and $0$ if $znotin E$. We can see that $X+Y$ is the number of $xin A$ such that $xin B$. Hence $X+Y=|B|=5$. We choose $5$-set randomly. So there's $binom{8}{5}$ possibilities of $B$. Among them, exactly $binom{7}{4}$ sets contain $x$.
$endgroup$
– Song
Dec 17 '18 at 5:51
add a comment |
$begingroup$
For $A$ any set, the notation $|A|$ denotes the cardinality of $A$, i.e. the number of elements the set $A$ contains.
answered Dec 17 '18 at 5:17
zoidbergzoidberg
1,080113
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As is pointed out, $|A|$ denotes the cardinality (the number of elements) of $A$. Thus we may write
$$
X = sum_{x=a,b,c,d} 1_{x in B},$$ and $$Y = sum_{x=e,f,g,h}1_{xin B}$$ where $1_{zin E}$ denotes the indicator function (taking values in ${0,1}$ whose value is $1$ if and only if the $zin E$.) We can immediately observe that
$$
X+Y =|B| = 5.
$$ This readily implies that $X$ and $Y$ are dependent. And since expectation operation is linear, we have
$$
E[X] = sum_{x=a,b,c,d} E[1_{x in B}] = sum_{x=a,b,c,d} P(x in B).
$$ For arbitrary $x$, it holds that
$$
P(xin B) = frac{binom{7}{4}}{binom{5}{8}} = frac{5}{8}.
$$Hence, this gives
$$
E[X] = 4cdotfrac{5}{8} = frac{5}{2},
$$ and $$
E[Y] =frac{5}{2}$$ also follows.
answered Dec 17 '18 at 5:34
SongSong
14.6k1635
$endgroup$
$begingroup$
I'm a little lost on the 1 z is an element in E part. What does that mean the indicator function taking values 0,1? Also, how is X+Y=|B|? Shouldn't it be equal to A because A has {a.b.c.d.e.f.g}? And where is 5Choose8 coming from?
$endgroup$
– Toby
Dec 17 '18 at 5:46
1
$begingroup$
I used $z$ and $E$ as generic variables. The Indicator function $1_{zin E}$, also called characteristic function, is a $0$-$1$ function which has value $1$ if $zin E$ and $0$ if $znotin E$. We can see that $X+Y$ is the number of $xin A$ such that $xin B$. Hence $X+Y=|B|=5$. We choose $5$-set randomly. So there's $binom{8}{5}$ possibilities of $B$. Among them, exactly $binom{7}{4}$ sets contain $x$.
$endgroup$
– Song
Dec 17 '18 at 5:51
add a comment |
$begingroup$
As is pointed out, $|A|$ denotes the cardinality (the number of elements) of $A$. Thus we may write
$$
X = sum_{x=a,b,c,d} 1_{x in B},$$ and $$Y = sum_{x=e,f,g,h}1_{xin B}$$ where $1_{zin E}$ denotes the indicator function (taking values in ${0,1}$ whose value is $1$ if and only if the $zin E$.) We can immediately observe that
$$
X+Y =|B| = 5.
$$ This readily implies that $X$ and $Y$ are dependent. And since expectation operation is linear, we have
$$
E[X] = sum_{x=a,b,c,d} E[1_{x in B}] = sum_{x=a,b,c,d} P(x in B).
$$ For arbitrary $x$, it holds that
$$
P(xin B) = frac{binom{7}{4}}{binom{5}{8}} = frac{5}{8}.
$$Hence, this gives
$$
E[X] = 4cdotfrac{5}{8} = frac{5}{2},
$$ and $$
E[Y] =frac{5}{2}$$ also follows.
answered Dec 17 '18 at 5:34
SongSong
14.6k1635
$endgroup$
$begingroup$
I'm a little lost on the 1 z is an element in E part. What does that mean the indicator function taking values 0,1? Also, how is X+Y=|B|? Shouldn't it be equal to A because A has {a.b.c.d.e.f.g}? And where is 5Choose8 coming from?
$endgroup$
– Toby
Dec 17 '18 at 5:46
1
$begingroup$
I used $z$ and $E$ as generic variables. The Indicator function $1_{zin E}$, also called characteristic function, is a $0$-$1$ function which has value $1$ if $zin E$ and $0$ if $znotin E$. We can see that $X+Y$ is the number of $xin A$ such that $xin B$. Hence $X+Y=|B|=5$. We choose $5$-set randomly. So there's $binom{8}{5}$ possibilities of $B$. Among them, exactly $binom{7}{4}$ sets contain $x$.
$endgroup$
– Song
Dec 17 '18 at 5:51
add a comment |
$begingroup$
As is pointed out, $|A|$ denotes the cardinality (the number of elements) of $A$. Thus we may write
$$
X = sum_{x=a,b,c,d} 1_{x in B},$$ and $$Y = sum_{x=e,f,g,h}1_{xin B}$$ where $1_{zin E}$ denotes the indicator function (taking values in ${0,1}$ whose value is $1$ if and only if the $zin E$.) We can immediately observe that
$$
X+Y =|B| = 5.
$$ This readily implies that $X$ and $Y$ are dependent. And since expectation operation is linear, we have
$$
E[X] = sum_{x=a,b,c,d} E[1_{x in B}] = sum_{x=a,b,c,d} P(x in B).
$$ For arbitrary $x$, it holds that
$$
P(xin B) = frac{binom{7}{4}}{binom{5}{8}} = frac{5}{8}.
$$Hence, this gives
$$
E[X] = 4cdotfrac{5}{8} = frac{5}{2},
$$ and $$
E[Y] =frac{5}{2}$$ also follows.
answered Dec 17 '18 at 5:34
SongSong
14.6k1635
$endgroup$
As is pointed out, $|A|$ denotes the cardinality (the number of elements) of $A$. Thus we may write
$$
X = sum_{x=a,b,c,d} 1_{x in B},$$ and $$Y = sum_{x=e,f,g,h}1_{xin B}$$ where $1_{zin E}$ denotes the indicator function (taking values in ${0,1}$ whose value is $1$ if and only if the $zin E$.) We can immediately observe that
$$
X+Y =|B| = 5.
$$ This readily implies that $X$ and $Y$ are dependent. And since expectation operation is linear, we have
$$
E[X] = sum_{x=a,b,c,d} E[1_{x in B}] = sum_{x=a,b,c,d} P(x in B).
$$ For arbitrary $x$, it holds that
$$
P(xin B) = frac{binom{7}{4}}{binom{5}{8}} = frac{5}{8}.
$$Hence, this gives
$$
E[X] = 4cdotfrac{5}{8} = frac{5}{2},
$$ and $$
E[Y] =frac{5}{2}$$ also follows.
answered Dec 17 '18 at 5:34
SongSong
14.6k1635
answered Dec 17 '18 at 5:34
SongSong
14.6k1635
answered Dec 17 '18 at 5:34
SongSong
14.6k1635
answered Dec 17 '18 at 5:34
SongSong
14.6k1635
14.6k1635
$begingroup$
I'm a little lost on the 1 z is an element in E part. What does that mean the indicator function taking values 0,1? Also, how is X+Y=|B|? Shouldn't it be equal to A because A has {a.b.c.d.e.f.g}? And where is 5Choose8 coming from?
$endgroup$
– Toby
Dec 17 '18 at 5:46
1
$begingroup$
I used $z$ and $E$ as generic variables. The Indicator function $1_{zin E}$, also called characteristic function, is a $0$-$1$ function which has value $1$ if $zin E$ and $0$ if $znotin E$. We can see that $X+Y$ is the number of $xin A$ such that $xin B$. Hence $X+Y=|B|=5$. We choose $5$-set randomly. So there's $binom{8}{5}$ possibilities of $B$. Among them, exactly $binom{7}{4}$ sets contain $x$.
$endgroup$
– Song
Dec 17 '18 at 5:51
add a comment |
$begingroup$
I'm a little lost on the 1 z is an element in E part. What does that mean the indicator function taking values 0,1? Also, how is X+Y=|B|? Shouldn't it be equal to A because A has {a.b.c.d.e.f.g}? And where is 5Choose8 coming from?
$endgroup$
– Toby
Dec 17 '18 at 5:46
1
$begingroup$
I used $z$ and $E$ as generic variables. The Indicator function $1_{zin E}$, also called characteristic function, is a $0$-$1$ function which has value $1$ if $zin E$ and $0$ if $znotin E$. We can see that $X+Y$ is the number of $xin A$ such that $xin B$. Hence $X+Y=|B|=5$. We choose $5$-set randomly. So there's $binom{8}{5}$ possibilities of $B$. Among them, exactly $binom{7}{4}$ sets contain $x$.
$endgroup$
– Song
Dec 17 '18 at 5:51
$begingroup$
I'm a little lost on the 1 z is an element in E part. What does that mean the indicator function taking values 0,1? Also, how is X+Y=|B|? Shouldn't it be equal to A because A has {a.b.c.d.e.f.g}? And where is 5Choose8 coming from?
$endgroup$
– Toby
Dec 17 '18 at 5:46
$begingroup$
I'm a little lost on the 1 z is an element in E part. What does that mean the indicator function taking values 0,1? Also, how is X+Y=|B|? Shouldn't it be equal to A because A has {a.b.c.d.e.f.g}? And where is 5Choose8 coming from?
$endgroup$
– Toby
Dec 17 '18 at 5:46
1
1
$begingroup$
I used $z$ and $E$ as generic variables. The Indicator function $1_{zin E}$, also called characteristic function, is a $0$-$1$ function which has value $1$ if $zin E$ and $0$ if $znotin E$. We can see that $X+Y$ is the number of $xin A$ such that $xin B$. Hence $X+Y=|B|=5$. We choose $5$-set randomly. So there's $binom{8}{5}$ possibilities of $B$. Among them, exactly $binom{7}{4}$ sets contain $x$.
$endgroup$
– Song
Dec 17 '18 at 5:51
$begingroup$
I used $z$ and $E$ as generic variables. The Indicator function $1_{zin E}$, also called characteristic function, is a $0$-$1$ function which has value $1$ if $zin E$ and $0$ if $znotin E$. We can see that $X+Y$ is the number of $xin A$ such that $xin B$. Hence $X+Y=|B|=5$. We choose $5$-set randomly. So there's $binom{8}{5}$ possibilities of $B$. Among them, exactly $binom{7}{4}$ sets contain $x$.
$endgroup$
– Song
Dec 17 '18 at 5:51
add a comment |
$begingroup$
For $A$ any set, the notation $|A|$ denotes the cardinality of $A$, i.e. the number of elements the set $A$ contains.
answered Dec 17 '18 at 5:17
zoidbergzoidberg
1,080113
$endgroup$
add a comment |
$begingroup$
For $A$ any set, the notation $|A|$ denotes the cardinality of $A$, i.e. the number of elements the set $A$ contains.
answered Dec 17 '18 at 5:17
zoidbergzoidberg
1,080113
$endgroup$
add a comment |
$begingroup$
For $A$ any set, the notation $|A|$ denotes the cardinality of $A$, i.e. the number of elements the set $A$ contains.
answered Dec 17 '18 at 5:17
zoidbergzoidberg
1,080113
$endgroup$
For $A$ any set, the notation $|A|$ denotes the cardinality of $A$, i.e. the number of elements the set $A$ contains.
answered Dec 17 '18 at 5:17
zoidbergzoidberg
1,080113
answered Dec 17 '18 at 5:17
zoidbergzoidberg
1,080113
answered Dec 17 '18 at 5:17
zoidbergzoidberg
1,080113
answered Dec 17 '18 at 5:17
zoidbergzoidberg
1,080113
1,080113
add a comment |
add a comment |
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