Mixing
My goal is to calculate $operatorname{Cov}(X,Y)$, but I'm struggling to calculate $E[XY]$
$begingroup$
Regarding the following problem:
A fair coin is tossed 5 times. Let $X$ be the number of heads in all $5$ tosses, and $Y$ be the number of heads in the first $4$ tosses. What is $operatorname{Cov}(X,Y)$?
My attempt:
I know that I should calculate the following:
$$ operatorname{Cov}(X,Y) = E[XY]-E[X]E[Y]$$
Well, the right flank is pretty easy: $E[X]=2.5, E[Y]=2 $
But what about $E[XY]$?
I searched online and saw that:
$$ E[XY]=sum_{x}sum_{y}xyP(X=x, Y=y) $$
But any attempt to imply that in the problem only led me for further more confusion.
Thanks in advance.
probability covariance means
edited Jan 22 at 19:59
StubbornAtom
6,16811339
asked Jan 22 at 16:29
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
Regarding the following problem:
A fair coin is tossed 5 times. Let $X$ be the number of heads in all $5$ tosses, and $Y$ be the number of heads in the first $4$ tosses. What is $operatorname{Cov}(X,Y)$?
My attempt:
I know that I should calculate the following:
$$ operatorname{Cov}(X,Y) = E[XY]-E[X]E[Y]$$
Well, the right flank is pretty easy: $E[X]=2.5, E[Y]=2 $
But what about $E[XY]$?
I searched online and saw that:
$$ E[XY]=sum_{x}sum_{y}xyP(X=x, Y=y) $$
But any attempt to imply that in the problem only led me for further more confusion.
Thanks in advance.
probability covariance means
edited Jan 22 at 19:59
StubbornAtom
6,16811339
asked Jan 22 at 16:29
superuser123superuser123
48628
$endgroup$
3
$begingroup$
Use the fact that $X=Y + Z$ where $Z$ is independent of $Y$ and $mathbb{P}(Z=1)=mathbb{P}(Z=0)=1/2$, this is because $X$ and $Y$ can differ only due to the last toss
$endgroup$
– Hayk
Jan 22 at 16:37
add a comment |
$begingroup$
Regarding the following problem:
A fair coin is tossed 5 times. Let $X$ be the number of heads in all $5$ tosses, and $Y$ be the number of heads in the first $4$ tosses. What is $operatorname{Cov}(X,Y)$?
My attempt:
I know that I should calculate the following:
$$ operatorname{Cov}(X,Y) = E[XY]-E[X]E[Y]$$
Well, the right flank is pretty easy: $E[X]=2.5, E[Y]=2 $
But what about $E[XY]$?
I searched online and saw that:
$$ E[XY]=sum_{x}sum_{y}xyP(X=x, Y=y) $$
But any attempt to imply that in the problem only led me for further more confusion.
Thanks in advance.
probability covariance means
edited Jan 22 at 19:59
StubbornAtom
6,16811339
asked Jan 22 at 16:29
superuser123superuser123
48628
$endgroup$
Regarding the following problem:
A fair coin is tossed 5 times. Let $X$ be the number of heads in all $5$ tosses, and $Y$ be the number of heads in the first $4$ tosses. What is $operatorname{Cov}(X,Y)$?
My attempt:
I know that I should calculate the following:
$$ operatorname{Cov}(X,Y) = E[XY]-E[X]E[Y]$$
Well, the right flank is pretty easy: $E[X]=2.5, E[Y]=2 $
But what about $E[XY]$?
I searched online and saw that:
$$ E[XY]=sum_{x}sum_{y}xyP(X=x, Y=y) $$
But any attempt to imply that in the problem only led me for further more confusion.
Thanks in advance.
probability covariance means
probability covariance means
edited Jan 22 at 19:59
StubbornAtom
6,16811339
asked Jan 22 at 16:29
superuser123superuser123
48628
edited Jan 22 at 19:59
StubbornAtom
6,16811339
asked Jan 22 at 16:29
superuser123superuser123
48628
edited Jan 22 at 19:59
StubbornAtom
6,16811339
edited Jan 22 at 19:59
StubbornAtom
6,16811339
edited Jan 22 at 19:59
StubbornAtom
6,16811339
6,16811339
asked Jan 22 at 16:29
superuser123superuser123
48628
asked Jan 22 at 16:29
superuser123superuser123
48628
asked Jan 22 at 16:29
superuser123superuser123
48628
48628
3
$begingroup$
Use the fact that $X=Y + Z$ where $Z$ is independent of $Y$ and $mathbb{P}(Z=1)=mathbb{P}(Z=0)=1/2$, this is because $X$ and $Y$ can differ only due to the last toss
$endgroup$
– Hayk
Jan 22 at 16:37
add a comment |
3
$begingroup$
Use the fact that $X=Y + Z$ where $Z$ is independent of $Y$ and $mathbb{P}(Z=1)=mathbb{P}(Z=0)=1/2$, this is because $X$ and $Y$ can differ only due to the last toss
$endgroup$
– Hayk
Jan 22 at 16:37
3
3
$begingroup$
Use the fact that $X=Y + Z$ where $Z$ is independent of $Y$ and $mathbb{P}(Z=1)=mathbb{P}(Z=0)=1/2$, this is because $X$ and $Y$ can differ only due to the last toss
$endgroup$
– Hayk
Jan 22 at 16:37
$begingroup$
Use the fact that $X=Y + Z$ where $Z$ is independent of $Y$ and $mathbb{P}(Z=1)=mathbb{P}(Z=0)=1/2$, this is because $X$ and $Y$ can differ only due to the last toss
$endgroup$
– Hayk
Jan 22 at 16:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here's a way to do it without messy arithmetic. Let $T_i$ be the result of the $i$'th toss ($0$ if tails, $1$ if heads). These are independent, with $X = T_1 + ldots + T_5$ and $Y = T_1 + ldots + T_4$. Then
$$text{Cov}(X,Y) = text{Cov}(Y+T_5, Y) = text{Cov}(Y,Y) + text{Cov}(T_5,Y) = text{Cov}(Y,Y) = text{Var}(Y)$$
Now since $T_1, ldots, T_4$ are independent,
$$text{Var}(Y) = text{Var}(T_1) + ldots + text{Var}(T_4) = 4 text{Var}(T_1)$$
and $text{Var}(T_1)$ is easy to find...
answered Jan 22 at 16:55
Robert IsraelRobert Israel
327k23216469
$endgroup$
$begingroup$
why does $Cov(T5,Y)=0$ ? thanks!
$endgroup$
– superuser123
Jan 22 at 17:00
1
$begingroup$
Because they are independent.
$endgroup$
– Robert Israel
Jan 22 at 17:10
add a comment |
$begingroup$
$X$ can assume the values $0,1,2,3,4,5$ and $Y$ can assume the values $0,1,2,3,4$. One of the terms in the sum, for example, is
$$5cdot 4cdot P(X=5,Y=4)=5cdot 4cdot frac{1}{2^{5}} $$
The rest of the terms are computed similarly. Note that you'll want to use
$$P(X=x,Y=y)=P(Y=y)cdot P(X=x|Y=y)$$
A helpful observation: $P(X=x,Y=y)=0$ if $x<y$ (since we cannot have more heads in the first $4$ tosses than in all $5$), so many of the terms in the sum are $0$.
answered Jan 22 at 16:37
pwerthpwerth
3,265417
$endgroup$
$begingroup$
If I sum it up it turns out to be $ frac{70}{32}$ which is incorrect, what did I miss here?
$endgroup$
– superuser123
Jan 22 at 17:27
$begingroup$
That is also what I'm getting for the sum. What makes you say it's incorrect?
$endgroup$
– pwerth
Jan 22 at 17:30
$begingroup$
$Cov(X,Y)=1$ according to the answers, and it doesn't come up like that, unless I'm missing something
$endgroup$
– superuser123
Jan 22 at 17:44
1
$begingroup$
My apologies, the sum is not equal to $frac{70}{32}$. I believe you assumed each outcome to have the same probability of $frac{1}{2^{5}}$ which is not correct. This is true for the example I gave in my answer because there is only $1$ outcome of the possible $2^{5}=32$ in which all $5$ flips are heads. But there are multiple outcomes where, say, we have $X=2,Y=1$, since the first "heads" can appear at any of the first $4$ flips.
$endgroup$
– pwerth
Jan 22 at 17:52
1
$begingroup$
The sum should work out to $frac{192}{32}=6$, which means the covariance is indeed $6-(2)(2.5)=1$
$endgroup$
– pwerth
Jan 22 at 17:54
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's a way to do it without messy arithmetic. Let $T_i$ be the result of the $i$'th toss ($0$ if tails, $1$ if heads). These are independent, with $X = T_1 + ldots + T_5$ and $Y = T_1 + ldots + T_4$. Then
$$text{Cov}(X,Y) = text{Cov}(Y+T_5, Y) = text{Cov}(Y,Y) + text{Cov}(T_5,Y) = text{Cov}(Y,Y) = text{Var}(Y)$$
Now since $T_1, ldots, T_4$ are independent,
$$text{Var}(Y) = text{Var}(T_1) + ldots + text{Var}(T_4) = 4 text{Var}(T_1)$$
and $text{Var}(T_1)$ is easy to find...
answered Jan 22 at 16:55
Robert IsraelRobert Israel
327k23216469
$endgroup$
$begingroup$
why does $Cov(T5,Y)=0$ ? thanks!
$endgroup$
– superuser123
Jan 22 at 17:00
1
$begingroup$
Because they are independent.
$endgroup$
– Robert Israel
Jan 22 at 17:10
add a comment |
$begingroup$
Here's a way to do it without messy arithmetic. Let $T_i$ be the result of the $i$'th toss ($0$ if tails, $1$ if heads). These are independent, with $X = T_1 + ldots + T_5$ and $Y = T_1 + ldots + T_4$. Then
$$text{Cov}(X,Y) = text{Cov}(Y+T_5, Y) = text{Cov}(Y,Y) + text{Cov}(T_5,Y) = text{Cov}(Y,Y) = text{Var}(Y)$$
Now since $T_1, ldots, T_4$ are independent,
$$text{Var}(Y) = text{Var}(T_1) + ldots + text{Var}(T_4) = 4 text{Var}(T_1)$$
and $text{Var}(T_1)$ is easy to find...
answered Jan 22 at 16:55
Robert IsraelRobert Israel
327k23216469
$endgroup$
$begingroup$
why does $Cov(T5,Y)=0$ ? thanks!
$endgroup$
– superuser123
Jan 22 at 17:00
1
$begingroup$
Because they are independent.
$endgroup$
– Robert Israel
Jan 22 at 17:10
add a comment |
$begingroup$
Here's a way to do it without messy arithmetic. Let $T_i$ be the result of the $i$'th toss ($0$ if tails, $1$ if heads). These are independent, with $X = T_1 + ldots + T_5$ and $Y = T_1 + ldots + T_4$. Then
$$text{Cov}(X,Y) = text{Cov}(Y+T_5, Y) = text{Cov}(Y,Y) + text{Cov}(T_5,Y) = text{Cov}(Y,Y) = text{Var}(Y)$$
Now since $T_1, ldots, T_4$ are independent,
$$text{Var}(Y) = text{Var}(T_1) + ldots + text{Var}(T_4) = 4 text{Var}(T_1)$$
and $text{Var}(T_1)$ is easy to find...
answered Jan 22 at 16:55
Robert IsraelRobert Israel
327k23216469
$endgroup$
Here's a way to do it without messy arithmetic. Let $T_i$ be the result of the $i$'th toss ($0$ if tails, $1$ if heads). These are independent, with $X = T_1 + ldots + T_5$ and $Y = T_1 + ldots + T_4$. Then
$$text{Cov}(X,Y) = text{Cov}(Y+T_5, Y) = text{Cov}(Y,Y) + text{Cov}(T_5,Y) = text{Cov}(Y,Y) = text{Var}(Y)$$
Now since $T_1, ldots, T_4$ are independent,
$$text{Var}(Y) = text{Var}(T_1) + ldots + text{Var}(T_4) = 4 text{Var}(T_1)$$
and $text{Var}(T_1)$ is easy to find...
answered Jan 22 at 16:55
Robert IsraelRobert Israel
327k23216469
answered Jan 22 at 16:55
Robert IsraelRobert Israel
327k23216469
answered Jan 22 at 16:55
Robert IsraelRobert Israel
327k23216469
answered Jan 22 at 16:55
Robert IsraelRobert Israel
327k23216469
327k23216469
$begingroup$
why does $Cov(T5,Y)=0$ ? thanks!
$endgroup$
– superuser123
Jan 22 at 17:00
1
$begingroup$
Because they are independent.
$endgroup$
– Robert Israel
Jan 22 at 17:10
add a comment |
$begingroup$
why does $Cov(T5,Y)=0$ ? thanks!
$endgroup$
– superuser123
Jan 22 at 17:00
1
$begingroup$
Because they are independent.
$endgroup$
– Robert Israel
Jan 22 at 17:10
$begingroup$
why does $Cov(T5,Y)=0$ ? thanks!
$endgroup$
– superuser123
Jan 22 at 17:00
$begingroup$
why does $Cov(T5,Y)=0$ ? thanks!
$endgroup$
– superuser123
Jan 22 at 17:00
1
1
$begingroup$
Because they are independent.
$endgroup$
– Robert Israel
Jan 22 at 17:10
$begingroup$
Because they are independent.
$endgroup$
– Robert Israel
Jan 22 at 17:10
add a comment |
$begingroup$
$X$ can assume the values $0,1,2,3,4,5$ and $Y$ can assume the values $0,1,2,3,4$. One of the terms in the sum, for example, is
$$5cdot 4cdot P(X=5,Y=4)=5cdot 4cdot frac{1}{2^{5}} $$
The rest of the terms are computed similarly. Note that you'll want to use
$$P(X=x,Y=y)=P(Y=y)cdot P(X=x|Y=y)$$
A helpful observation: $P(X=x,Y=y)=0$ if $x<y$ (since we cannot have more heads in the first $4$ tosses than in all $5$), so many of the terms in the sum are $0$.
answered Jan 22 at 16:37
pwerthpwerth
3,265417
$endgroup$
$begingroup$
If I sum it up it turns out to be $ frac{70}{32}$ which is incorrect, what did I miss here?
$endgroup$
– superuser123
Jan 22 at 17:27
$begingroup$
That is also what I'm getting for the sum. What makes you say it's incorrect?
$endgroup$
– pwerth
Jan 22 at 17:30
$begingroup$
$Cov(X,Y)=1$ according to the answers, and it doesn't come up like that, unless I'm missing something
$endgroup$
– superuser123
Jan 22 at 17:44
1
$begingroup$
My apologies, the sum is not equal to $frac{70}{32}$. I believe you assumed each outcome to have the same probability of $frac{1}{2^{5}}$ which is not correct. This is true for the example I gave in my answer because there is only $1$ outcome of the possible $2^{5}=32$ in which all $5$ flips are heads. But there are multiple outcomes where, say, we have $X=2,Y=1$, since the first "heads" can appear at any of the first $4$ flips.
$endgroup$
– pwerth
Jan 22 at 17:52
1
$begingroup$
The sum should work out to $frac{192}{32}=6$, which means the covariance is indeed $6-(2)(2.5)=1$
$endgroup$
– pwerth
Jan 22 at 17:54
add a comment |
$begingroup$
$X$ can assume the values $0,1,2,3,4,5$ and $Y$ can assume the values $0,1,2,3,4$. One of the terms in the sum, for example, is
$$5cdot 4cdot P(X=5,Y=4)=5cdot 4cdot frac{1}{2^{5}} $$
The rest of the terms are computed similarly. Note that you'll want to use
$$P(X=x,Y=y)=P(Y=y)cdot P(X=x|Y=y)$$
A helpful observation: $P(X=x,Y=y)=0$ if $x<y$ (since we cannot have more heads in the first $4$ tosses than in all $5$), so many of the terms in the sum are $0$.
answered Jan 22 at 16:37
pwerthpwerth
3,265417
$endgroup$
$begingroup$
If I sum it up it turns out to be $ frac{70}{32}$ which is incorrect, what did I miss here?
$endgroup$
– superuser123
Jan 22 at 17:27
$begingroup$
That is also what I'm getting for the sum. What makes you say it's incorrect?
$endgroup$
– pwerth
Jan 22 at 17:30
$begingroup$
$Cov(X,Y)=1$ according to the answers, and it doesn't come up like that, unless I'm missing something
$endgroup$
– superuser123
Jan 22 at 17:44
1
$begingroup$
My apologies, the sum is not equal to $frac{70}{32}$. I believe you assumed each outcome to have the same probability of $frac{1}{2^{5}}$ which is not correct. This is true for the example I gave in my answer because there is only $1$ outcome of the possible $2^{5}=32$ in which all $5$ flips are heads. But there are multiple outcomes where, say, we have $X=2,Y=1$, since the first "heads" can appear at any of the first $4$ flips.
$endgroup$
– pwerth
Jan 22 at 17:52
1
$begingroup$
The sum should work out to $frac{192}{32}=6$, which means the covariance is indeed $6-(2)(2.5)=1$
$endgroup$
– pwerth
Jan 22 at 17:54
add a comment |
$begingroup$
$X$ can assume the values $0,1,2,3,4,5$ and $Y$ can assume the values $0,1,2,3,4$. One of the terms in the sum, for example, is
$$5cdot 4cdot P(X=5,Y=4)=5cdot 4cdot frac{1}{2^{5}} $$
The rest of the terms are computed similarly. Note that you'll want to use
$$P(X=x,Y=y)=P(Y=y)cdot P(X=x|Y=y)$$
A helpful observation: $P(X=x,Y=y)=0$ if $x<y$ (since we cannot have more heads in the first $4$ tosses than in all $5$), so many of the terms in the sum are $0$.
answered Jan 22 at 16:37
pwerthpwerth
3,265417
$endgroup$
$X$ can assume the values $0,1,2,3,4,5$ and $Y$ can assume the values $0,1,2,3,4$. One of the terms in the sum, for example, is
$$5cdot 4cdot P(X=5,Y=4)=5cdot 4cdot frac{1}{2^{5}} $$
The rest of the terms are computed similarly. Note that you'll want to use
$$P(X=x,Y=y)=P(Y=y)cdot P(X=x|Y=y)$$
A helpful observation: $P(X=x,Y=y)=0$ if $x<y$ (since we cannot have more heads in the first $4$ tosses than in all $5$), so many of the terms in the sum are $0$.
answered Jan 22 at 16:37
pwerthpwerth
3,265417
answered Jan 22 at 16:37
pwerthpwerth
3,265417
answered Jan 22 at 16:37
pwerthpwerth
3,265417
answered Jan 22 at 16:37
pwerthpwerth
3,265417
3,265417
$begingroup$
If I sum it up it turns out to be $ frac{70}{32}$ which is incorrect, what did I miss here?
$endgroup$
– superuser123
Jan 22 at 17:27
$begingroup$
That is also what I'm getting for the sum. What makes you say it's incorrect?
$endgroup$
– pwerth
Jan 22 at 17:30
$begingroup$
$Cov(X,Y)=1$ according to the answers, and it doesn't come up like that, unless I'm missing something
$endgroup$
– superuser123
Jan 22 at 17:44
1
$begingroup$
My apologies, the sum is not equal to $frac{70}{32}$. I believe you assumed each outcome to have the same probability of $frac{1}{2^{5}}$ which is not correct. This is true for the example I gave in my answer because there is only $1$ outcome of the possible $2^{5}=32$ in which all $5$ flips are heads. But there are multiple outcomes where, say, we have $X=2,Y=1$, since the first "heads" can appear at any of the first $4$ flips.
$endgroup$
– pwerth
Jan 22 at 17:52
1
$begingroup$
The sum should work out to $frac{192}{32}=6$, which means the covariance is indeed $6-(2)(2.5)=1$
$endgroup$
– pwerth
Jan 22 at 17:54
add a comment |
$begingroup$
If I sum it up it turns out to be $ frac{70}{32}$ which is incorrect, what did I miss here?
$endgroup$
– superuser123
Jan 22 at 17:27
$begingroup$
That is also what I'm getting for the sum. What makes you say it's incorrect?
$endgroup$
– pwerth
Jan 22 at 17:30
$begingroup$
$Cov(X,Y)=1$ according to the answers, and it doesn't come up like that, unless I'm missing something
$endgroup$
– superuser123
Jan 22 at 17:44
1
$begingroup$
My apologies, the sum is not equal to $frac{70}{32}$. I believe you assumed each outcome to have the same probability of $frac{1}{2^{5}}$ which is not correct. This is true for the example I gave in my answer because there is only $1$ outcome of the possible $2^{5}=32$ in which all $5$ flips are heads. But there are multiple outcomes where, say, we have $X=2,Y=1$, since the first "heads" can appear at any of the first $4$ flips.
$endgroup$
– pwerth
Jan 22 at 17:52
1
$begingroup$
The sum should work out to $frac{192}{32}=6$, which means the covariance is indeed $6-(2)(2.5)=1$
$endgroup$
– pwerth
Jan 22 at 17:54
$begingroup$
If I sum it up it turns out to be $ frac{70}{32}$ which is incorrect, what did I miss here?
$endgroup$
– superuser123
Jan 22 at 17:27
$begingroup$
If I sum it up it turns out to be $ frac{70}{32}$ which is incorrect, what did I miss here?
$endgroup$
– superuser123
Jan 22 at 17:27
$begingroup$
That is also what I'm getting for the sum. What makes you say it's incorrect?
$endgroup$
– pwerth
Jan 22 at 17:30
$begingroup$
That is also what I'm getting for the sum. What makes you say it's incorrect?
$endgroup$
– pwerth
Jan 22 at 17:30
$begingroup$
$Cov(X,Y)=1$ according to the answers, and it doesn't come up like that, unless I'm missing something
$endgroup$
– superuser123
Jan 22 at 17:44
$begingroup$
$Cov(X,Y)=1$ according to the answers, and it doesn't come up like that, unless I'm missing something
$endgroup$
– superuser123
Jan 22 at 17:44
1
1
$begingroup$
My apologies, the sum is not equal to $frac{70}{32}$. I believe you assumed each outcome to have the same probability of $frac{1}{2^{5}}$ which is not correct. This is true for the example I gave in my answer because there is only $1$ outcome of the possible $2^{5}=32$ in which all $5$ flips are heads. But there are multiple outcomes where, say, we have $X=2,Y=1$, since the first "heads" can appear at any of the first $4$ flips.
$endgroup$
– pwerth
Jan 22 at 17:52
$begingroup$
My apologies, the sum is not equal to $frac{70}{32}$. I believe you assumed each outcome to have the same probability of $frac{1}{2^{5}}$ which is not correct. This is true for the example I gave in my answer because there is only $1$ outcome of the possible $2^{5}=32$ in which all $5$ flips are heads. But there are multiple outcomes where, say, we have $X=2,Y=1$, since the first "heads" can appear at any of the first $4$ flips.
$endgroup$
– pwerth
Jan 22 at 17:52
1
1
$begingroup$
The sum should work out to $frac{192}{32}=6$, which means the covariance is indeed $6-(2)(2.5)=1$
$endgroup$
– pwerth
Jan 22 at 17:54
$begingroup$
The sum should work out to $frac{192}{32}=6$, which means the covariance is indeed $6-(2)(2.5)=1$
$endgroup$
– pwerth
Jan 22 at 17:54
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$begingroup$
Use the fact that $X=Y + Z$ where $Z$ is independent of $Y$ and $mathbb{P}(Z=1)=mathbb{P}(Z=0)=1/2$, this is because $X$ and $Y$ can differ only due to the last toss
$endgroup$
– Hayk
Jan 22 at 16:37