Find the standard deviation of $|X−Y|$
$begingroup$
Let $X$ and $Y$ be independent random variables with a Bernoulli Distribution $Ber(1/3)$. Find the standard deviation of $|X−Y|$.
The Standard Deviation in the square root of the variance. For a $Ber(1/3)$ the $Var(X)=Var(Y)=1/3(1-1/3)=2/9$, now how can I calculate $sigma=sqrt{Var(|X−Y|)}$? Because my idea was to subtract the two $Var$ but then the result will be $0$, but it should be $frac{2sqrt{5}}{9}$, how can I solve it?
probability statistics
asked Jan 25 at 16:06
FTACFTAC
2649
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be independent random variables with a Bernoulli Distribution $Ber(1/3)$. Find the standard deviation of $|X−Y|$.
The Standard Deviation in the square root of the variance. For a $Ber(1/3)$ the $Var(X)=Var(Y)=1/3(1-1/3)=2/9$, now how can I calculate $sigma=sqrt{Var(|X−Y|)}$? Because my idea was to subtract the two $Var$ but then the result will be $0$, but it should be $frac{2sqrt{5}}{9}$, how can I solve it?
probability statistics
asked Jan 25 at 16:06
FTACFTAC
2649
$endgroup$
3
$begingroup$
$Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
$endgroup$
– Robert Israel
Jan 25 at 16:18
$begingroup$
@RobertIsrael yes it's bernoulli, which four cases?
$endgroup$
– FTAC
Jan 25 at 16:20
add a comment |
$begingroup$
Let $X$ and $Y$ be independent random variables with a Bernoulli Distribution $Ber(1/3)$. Find the standard deviation of $|X−Y|$.
The Standard Deviation in the square root of the variance. For a $Ber(1/3)$ the $Var(X)=Var(Y)=1/3(1-1/3)=2/9$, now how can I calculate $sigma=sqrt{Var(|X−Y|)}$? Because my idea was to subtract the two $Var$ but then the result will be $0$, but it should be $frac{2sqrt{5}}{9}$, how can I solve it?
probability statistics
asked Jan 25 at 16:06
FTACFTAC
2649
$endgroup$
Let $X$ and $Y$ be independent random variables with a Bernoulli Distribution $Ber(1/3)$. Find the standard deviation of $|X−Y|$.
The Standard Deviation in the square root of the variance. For a $Ber(1/3)$ the $Var(X)=Var(Y)=1/3(1-1/3)=2/9$, now how can I calculate $sigma=sqrt{Var(|X−Y|)}$? Because my idea was to subtract the two $Var$ but then the result will be $0$, but it should be $frac{2sqrt{5}}{9}$, how can I solve it?
probability statistics
probability statistics
asked Jan 25 at 16:06
FTACFTAC
2649
asked Jan 25 at 16:06
FTACFTAC
2649
edited Jan 25 at 16:19
FTAC
asked Jan 25 at 16:06
FTACFTAC
2649
asked Jan 25 at 16:06
FTACFTAC
2649
asked Jan 25 at 16:06
FTACFTAC
2649
2649
3
$begingroup$
$Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
$endgroup$
– Robert Israel
Jan 25 at 16:18
$begingroup$
@RobertIsrael yes it's bernoulli, which four cases?
$endgroup$
– FTAC
Jan 25 at 16:20
add a comment |
3
$begingroup$
$Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
$endgroup$
– Robert Israel
Jan 25 at 16:18
$begingroup$
@RobertIsrael yes it's bernoulli, which four cases?
$endgroup$
– FTAC
Jan 25 at 16:20
3
3
$begingroup$
$Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
$endgroup$
– Robert Israel
Jan 25 at 16:18
$begingroup$
$Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
$endgroup$
– Robert Israel
Jan 25 at 16:18
$begingroup$
@RobertIsrael yes it's bernoulli, which four cases?
$endgroup$
– FTAC
Jan 25 at 16:20
$begingroup$
@RobertIsrael yes it's bernoulli, which four cases?
$endgroup$
– FTAC
Jan 25 at 16:20
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
begin{equation}
|X-Y|= begin{cases}
1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
end{cases}
end{equation}
So $|X-Y|$ is another Bernoulli random variable and
$Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$
and then
$Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
as required.
answered Jan 25 at 16:55
amator2357amator2357
729
$endgroup$
$begingroup$
thanks for the answer, what does Pr mean?
$endgroup$
– FTAC
Jan 25 at 17:00
2
$begingroup$
@FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
$endgroup$
– callculus
Jan 25 at 17:25
add a comment |
$begingroup$
Hints:
$X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.
In the first two cases their difference would be $1$ while in the last two it would be $0$.
So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root
answered Jan 25 at 16:37
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
In general, let $X,Ysim B(p)$. Refer to the table:
$$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
hline
0&q&0&q&0&q^2&0&q^2\
0&q&1&p&1&qp&1&qp\
1&p&0&q&1&pq&1&pq\
1&p&1&p&0&p^2&0&p^2\
end{array}\
text{1-method}: |X-Y|sim B(2pq); \
Var(|X-Y|)=2pqcdot (1-2pq).\
text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
2pq-[2pq]^2 =2pqcdot (1-2pq).$$
answered Jan 25 at 17:15
farruhotafarruhota
20.8k2741
$endgroup$
add a comment |
$begingroup$
The easiest thing is to use a table to obtain the distribution.
The pdf of $|X-Y|$
$$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$
Now you can use the well known formulas to obtain $Var(|X-Y|)$
$E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$
$E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
p(x,y)$$Var(Z)=E(Z^2)-E^2(Z)$
answered Jan 25 at 17:03
callculuscallculus
18.3k31427
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
begin{equation}
|X-Y|= begin{cases}
1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
end{cases}
end{equation}
So $|X-Y|$ is another Bernoulli random variable and
$Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$
and then
$Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
as required.
answered Jan 25 at 16:55
amator2357amator2357
729
$endgroup$
$begingroup$
thanks for the answer, what does Pr mean?
$endgroup$
– FTAC
Jan 25 at 17:00
2
$begingroup$
@FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
$endgroup$
– callculus
Jan 25 at 17:25
add a comment |
$begingroup$
begin{equation}
|X-Y|= begin{cases}
1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
end{cases}
end{equation}
So $|X-Y|$ is another Bernoulli random variable and
$Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$
and then
$Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
as required.
answered Jan 25 at 16:55
amator2357amator2357
729
$endgroup$
$begingroup$
thanks for the answer, what does Pr mean?
$endgroup$
– FTAC
Jan 25 at 17:00
2
$begingroup$
@FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
$endgroup$
– callculus
Jan 25 at 17:25
add a comment |
$begingroup$
begin{equation}
|X-Y|= begin{cases}
1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
end{cases}
end{equation}
So $|X-Y|$ is another Bernoulli random variable and
$Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$
and then
$Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
as required.
answered Jan 25 at 16:55
amator2357amator2357
729
$endgroup$
begin{equation}
|X-Y|= begin{cases}
1 & text{iff} quad (X=1 quad wedgequad Y=0)quad vee quad (X=0 quad wedgequad Y=1) \
0 & text{iff} quad (X=1 quad wedgequad Y=1)quad vee quad (X=0 quad wedgequad Y=0)
end{cases}
end{equation}
So $|X-Y|$ is another Bernoulli random variable and
$Pr(|X-Y|=1)=Pr(X=1)Pr(Y=0)+Pr(X=0)Pr(Y=1)=(1/3)(2/3)+(2/3)(1/3)=4/9=p$
and then
$Var(|X-Y|)=(4/9)(5/9)=20/81 implies SD=sqrt{20/81}=(2sqrt{5})/9$,
as required.
answered Jan 25 at 16:55
amator2357amator2357
729
answered Jan 25 at 16:55
amator2357amator2357
729
answered Jan 25 at 16:55
amator2357amator2357
729
answered Jan 25 at 16:55
amator2357amator2357
729
729
$begingroup$
thanks for the answer, what does Pr mean?
$endgroup$
– FTAC
Jan 25 at 17:00
2
$begingroup$
@FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
$endgroup$
– callculus
Jan 25 at 17:25
add a comment |
$begingroup$
thanks for the answer, what does Pr mean?
$endgroup$
– FTAC
Jan 25 at 17:00
2
$begingroup$
@FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
$endgroup$
– callculus
Jan 25 at 17:25
$begingroup$
thanks for the answer, what does Pr mean?
$endgroup$
– FTAC
Jan 25 at 17:00
$begingroup$
thanks for the answer, what does Pr mean?
$endgroup$
– FTAC
Jan 25 at 17:00
2
2
$begingroup$
@FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
$endgroup$
– callculus
Jan 25 at 17:25
$begingroup$
@FabioTaccaliti The same as $P()$ (in statistical context). $P(X=0)$ or $Pr(X=0)$ is the probability that the random variable $X$ takes the value $0$.
$endgroup$
– callculus
Jan 25 at 17:25
add a comment |
$begingroup$
Hints:
$X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.
In the first two cases their difference would be $1$ while in the last two it would be $0$.
So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root
answered Jan 25 at 16:37
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
Hints:
$X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.
In the first two cases their difference would be $1$ while in the last two it would be $0$.
So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root
answered Jan 25 at 16:37
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
Hints:
$X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.
In the first two cases their difference would be $1$ while in the last two it would be $0$.
So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root
answered Jan 25 at 16:37
HenryHenry
101k481168
$endgroup$
Hints:
$X$ could be $1$ and $Y$ could be $0$, or the other way round, or both $1$, or both $0$.
In the first two cases their difference would be $1$ while in the last two it would be $0$.
So $|X-Y|$ is another Bernoulli random variable, and you need to find the probability it is $1$ and then calculate the variance and take its square root
answered Jan 25 at 16:37
HenryHenry
101k481168
answered Jan 25 at 16:37
HenryHenry
101k481168
answered Jan 25 at 16:37
HenryHenry
101k481168
answered Jan 25 at 16:37
HenryHenry
101k481168
101k481168
add a comment |
add a comment |
$begingroup$
In general, let $X,Ysim B(p)$. Refer to the table:
$$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
hline
0&q&0&q&0&q^2&0&q^2\
0&q&1&p&1&qp&1&qp\
1&p&0&q&1&pq&1&pq\
1&p&1&p&0&p^2&0&p^2\
end{array}\
text{1-method}: |X-Y|sim B(2pq); \
Var(|X-Y|)=2pqcdot (1-2pq).\
text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
2pq-[2pq]^2 =2pqcdot (1-2pq).$$
answered Jan 25 at 17:15
farruhotafarruhota
20.8k2741
$endgroup$
add a comment |
$begingroup$
In general, let $X,Ysim B(p)$. Refer to the table:
$$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
hline
0&q&0&q&0&q^2&0&q^2\
0&q&1&p&1&qp&1&qp\
1&p&0&q&1&pq&1&pq\
1&p&1&p&0&p^2&0&p^2\
end{array}\
text{1-method}: |X-Y|sim B(2pq); \
Var(|X-Y|)=2pqcdot (1-2pq).\
text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
2pq-[2pq]^2 =2pqcdot (1-2pq).$$
answered Jan 25 at 17:15
farruhotafarruhota
20.8k2741
$endgroup$
add a comment |
$begingroup$
In general, let $X,Ysim B(p)$. Refer to the table:
$$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
hline
0&q&0&q&0&q^2&0&q^2\
0&q&1&p&1&qp&1&qp\
1&p&0&q&1&pq&1&pq\
1&p&1&p&0&p^2&0&p^2\
end{array}\
text{1-method}: |X-Y|sim B(2pq); \
Var(|X-Y|)=2pqcdot (1-2pq).\
text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
2pq-[2pq]^2 =2pqcdot (1-2pq).$$
answered Jan 25 at 17:15
farruhotafarruhota
20.8k2741
$endgroup$
In general, let $X,Ysim B(p)$. Refer to the table:
$$begin{array}{c|c|c|c|c|c} X & P(X)&Y&P(Y)&|X-Y|&P(|X-Y|)&|X-Y|^2&P(|X-Y|^2)\
hline
0&q&0&q&0&q^2&0&q^2\
0&q&1&p&1&qp&1&qp\
1&p&0&q&1&pq&1&pq\
1&p&1&p&0&p^2&0&p^2\
end{array}\
text{1-method}: |X-Y|sim B(2pq); \
Var(|X-Y|)=2pqcdot (1-2pq).\
text{2-method}: Var(|X-Y|)=mathbb E(|X-Y|^2)-[mathbb E(|X-Y|)]^2=\
2pq-[2pq]^2 =2pqcdot (1-2pq).$$
answered Jan 25 at 17:15
farruhotafarruhota
20.8k2741
answered Jan 25 at 17:15
farruhotafarruhota
20.8k2741
answered Jan 25 at 17:15
farruhotafarruhota
20.8k2741
answered Jan 25 at 17:15
farruhotafarruhota
20.8k2741
20.8k2741
add a comment |
add a comment |
$begingroup$
The easiest thing is to use a table to obtain the distribution.
The pdf of $|X-Y|$
$$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$
Now you can use the well known formulas to obtain $Var(|X-Y|)$
$E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$
$E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
p(x,y)$$Var(Z)=E(Z^2)-E^2(Z)$
answered Jan 25 at 17:03
callculuscallculus
18.3k31427
$endgroup$
add a comment |
$begingroup$
The easiest thing is to use a table to obtain the distribution.
The pdf of $|X-Y|$
$$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$
Now you can use the well known formulas to obtain $Var(|X-Y|)$
$E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$
$E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
p(x,y)$$Var(Z)=E(Z^2)-E^2(Z)$
answered Jan 25 at 17:03
callculuscallculus
18.3k31427
$endgroup$
add a comment |
$begingroup$
The easiest thing is to use a table to obtain the distribution.
The pdf of $|X-Y|$
$$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$
Now you can use the well known formulas to obtain $Var(|X-Y|)$
$E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$
$E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
p(x,y)$$Var(Z)=E(Z^2)-E^2(Z)$
answered Jan 25 at 17:03
callculuscallculus
18.3k31427
$endgroup$
The easiest thing is to use a table to obtain the distribution.
The pdf of $|X-Y|$
$$begin{array}{|c|c|c|c|} hline X/Y & 0left(p=frac13right)&1left(p=frac23right) \ hline 0left(p=frac13right) & 0 &1 \ hline 1left(p=frac23right) & 1 &0 \ hline end{array}$$
Now you can use the well known formulas to obtain $Var(|X-Y|)$
$E(|X-Y|)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|cdot p(x,y)$
$E(|X-Y|^2)=sumlimits_{x=0}^{1}sumlimits_{y=0}^{1} |x-y|^2cdot
p(x,y)$$Var(Z)=E(Z^2)-E^2(Z)$
answered Jan 25 at 17:03
callculuscallculus
18.3k31427
answered Jan 25 at 17:03
callculuscallculus
18.3k31427
answered Jan 25 at 17:03
callculuscallculus
18.3k31427
answered Jan 25 at 17:03
callculuscallculus
18.3k31427
18.3k31427
add a comment |
add a comment |
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3
$begingroup$
$Ber$ = bernoulli? There are only $4$ outcomes to consider, so it's easy to do this from the definition.
$endgroup$
– Robert Israel
Jan 25 at 16:18
$begingroup$
@RobertIsrael yes it's bernoulli, which four cases?
$endgroup$
– FTAC
Jan 25 at 16:20