How to prove $P(Acup B) = P(A) + P(B) - P(Acap B)$
$begingroup$
Let $X$ and $Y$ any random variables. $A$ and $B$ are two events $in Omega$ (the sample space):
How can I prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ using the equation: $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],; A, B in Omega$ (my sample space)
I believe to prove this $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)]$
I will have to test for $X>Y$, $X<Y$ and $X=Y$ right?
And to use this equation to prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ I will have to use the idea of $A subset B, ; B subset A$ and $A = B,$ right?
probability
edited Jan 18 at 22:05
Andrés E. Caicedo
66.1k8160252
asked Jan 18 at 20:31
LauraLaura
3758
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ any random variables. $A$ and $B$ are two events $in Omega$ (the sample space):
How can I prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ using the equation: $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],; A, B in Omega$ (my sample space)
I believe to prove this $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)]$
I will have to test for $X>Y$, $X<Y$ and $X=Y$ right?
And to use this equation to prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ I will have to use the idea of $A subset B, ; B subset A$ and $A = B,$ right?
probability
edited Jan 18 at 22:05
Andrés E. Caicedo
66.1k8160252
asked Jan 18 at 20:31
LauraLaura
3758
$endgroup$
1
$begingroup$
you should porably explain your notations.
$endgroup$
– J.F
Jan 18 at 20:33
$begingroup$
@G.F Iedited the question. Thanks.
$endgroup$
– Laura
Jan 18 at 20:36
1
$begingroup$
I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
$endgroup$
– WaveX
Jan 18 at 20:37
1
$begingroup$
It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
$endgroup$
– JMoravitz
Jan 18 at 20:37
$begingroup$
@JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
$endgroup$
– Laura
Jan 18 at 20:41
add a comment |
$begingroup$
Let $X$ and $Y$ any random variables. $A$ and $B$ are two events $in Omega$ (the sample space):
How can I prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ using the equation: $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],; A, B in Omega$ (my sample space)
I believe to prove this $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)]$
I will have to test for $X>Y$, $X<Y$ and $X=Y$ right?
And to use this equation to prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ I will have to use the idea of $A subset B, ; B subset A$ and $A = B,$ right?
probability
edited Jan 18 at 22:05
Andrés E. Caicedo
66.1k8160252
asked Jan 18 at 20:31
LauraLaura
3758
$endgroup$
Let $X$ and $Y$ any random variables. $A$ and $B$ are two events $in Omega$ (the sample space):
How can I prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ using the equation: $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],; A, B in Omega$ (my sample space)
I believe to prove this $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)]$
I will have to test for $X>Y$, $X<Y$ and $X=Y$ right?
And to use this equation to prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ I will have to use the idea of $A subset B, ; B subset A$ and $A = B,$ right?
probability
probability
edited Jan 18 at 22:05
Andrés E. Caicedo
66.1k8160252
asked Jan 18 at 20:31
LauraLaura
3758
edited Jan 18 at 22:05
Andrés E. Caicedo
66.1k8160252
asked Jan 18 at 20:31
LauraLaura
3758
edited Jan 18 at 22:05
Andrés E. Caicedo
66.1k8160252
edited Jan 18 at 22:05
Andrés E. Caicedo
66.1k8160252
edited Jan 18 at 22:05
Andrés E. Caicedo
66.1k8160252
66.1k8160252
asked Jan 18 at 20:31
LauraLaura
3758
asked Jan 18 at 20:31
LauraLaura
3758
asked Jan 18 at 20:31
LauraLaura
3758
3758
1
$begingroup$
you should porably explain your notations.
$endgroup$
– J.F
Jan 18 at 20:33
$begingroup$
@G.F Iedited the question. Thanks.
$endgroup$
– Laura
Jan 18 at 20:36
1
$begingroup$
I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
$endgroup$
– WaveX
Jan 18 at 20:37
1
$begingroup$
It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
$endgroup$
– JMoravitz
Jan 18 at 20:37
$begingroup$
@JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
$endgroup$
– Laura
Jan 18 at 20:41
add a comment |
1
$begingroup$
you should porably explain your notations.
$endgroup$
– J.F
Jan 18 at 20:33
$begingroup$
@G.F Iedited the question. Thanks.
$endgroup$
– Laura
Jan 18 at 20:36
1
$begingroup$
I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
$endgroup$
– WaveX
Jan 18 at 20:37
1
$begingroup$
It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
$endgroup$
– JMoravitz
Jan 18 at 20:37
$begingroup$
@JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
$endgroup$
– Laura
Jan 18 at 20:41
1
1
$begingroup$
you should porably explain your notations.
$endgroup$
– J.F
Jan 18 at 20:33
$begingroup$
you should porably explain your notations.
$endgroup$
– J.F
Jan 18 at 20:33
$begingroup$
@G.F Iedited the question. Thanks.
$endgroup$
– Laura
Jan 18 at 20:36
$begingroup$
@G.F Iedited the question. Thanks.
$endgroup$
– Laura
Jan 18 at 20:36
1
1
$begingroup$
I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
$endgroup$
– WaveX
Jan 18 at 20:37
$begingroup$
I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
$endgroup$
– WaveX
Jan 18 at 20:37
1
1
$begingroup$
It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
$endgroup$
– JMoravitz
Jan 18 at 20:37
$begingroup$
It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
$endgroup$
– JMoravitz
Jan 18 at 20:37
$begingroup$
@JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
$endgroup$
– Laura
Jan 18 at 20:41
$begingroup$
@JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
$endgroup$
– Laura
Jan 18 at 20:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.
answered Jan 18 at 20:44
W-t-PW-t-P
1,954612
$endgroup$
$begingroup$
thanks @W-t-P! But how can I understand this concept of Indicator variable?
$endgroup$
– Laura
Jan 19 at 1:20
1
$begingroup$
@Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
$endgroup$
– WaveX
Jan 19 at 3:41
add a comment |
$begingroup$
If you need to prove that
$$
E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
$$
simply note that the relation is valid even before you take expectations:
$$
max(X,Y) = X + Y - min(X,Y),
$$
which is the same as
$$
X + Y = min(X,Y) + max(X,Y),
$$
which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).
answered Jan 18 at 22:50
grand_chatgrand_chat
20.5k11327
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
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votes
$begingroup$
If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.
answered Jan 18 at 20:44
W-t-PW-t-P
1,954612
$endgroup$
$begingroup$
thanks @W-t-P! But how can I understand this concept of Indicator variable?
$endgroup$
– Laura
Jan 19 at 1:20
1
$begingroup$
@Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
$endgroup$
– WaveX
Jan 19 at 3:41
add a comment |
$begingroup$
If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.
answered Jan 18 at 20:44
W-t-PW-t-P
1,954612
$endgroup$
$begingroup$
thanks @W-t-P! But how can I understand this concept of Indicator variable?
$endgroup$
– Laura
Jan 19 at 1:20
1
$begingroup$
@Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
$endgroup$
– WaveX
Jan 19 at 3:41
add a comment |
$begingroup$
If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.
answered Jan 18 at 20:44
W-t-PW-t-P
1,954612
$endgroup$
If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.
answered Jan 18 at 20:44
W-t-PW-t-P
1,954612
answered Jan 18 at 20:44
W-t-PW-t-P
1,954612
answered Jan 18 at 20:44
W-t-PW-t-P
1,954612
answered Jan 18 at 20:44
W-t-PW-t-P
1,954612
1,954612
$begingroup$
thanks @W-t-P! But how can I understand this concept of Indicator variable?
$endgroup$
– Laura
Jan 19 at 1:20
1
$begingroup$
@Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
$endgroup$
– WaveX
Jan 19 at 3:41
add a comment |
$begingroup$
thanks @W-t-P! But how can I understand this concept of Indicator variable?
$endgroup$
– Laura
Jan 19 at 1:20
1
$begingroup$
@Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
$endgroup$
– WaveX
Jan 19 at 3:41
$begingroup$
thanks @W-t-P! But how can I understand this concept of Indicator variable?
$endgroup$
– Laura
Jan 19 at 1:20
$begingroup$
thanks @W-t-P! But how can I understand this concept of Indicator variable?
$endgroup$
– Laura
Jan 19 at 1:20
1
1
$begingroup$
@Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
$endgroup$
– WaveX
Jan 19 at 3:41
$begingroup$
@Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
$endgroup$
– WaveX
Jan 19 at 3:41
add a comment |
$begingroup$
If you need to prove that
$$
E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
$$
simply note that the relation is valid even before you take expectations:
$$
max(X,Y) = X + Y - min(X,Y),
$$
which is the same as
$$
X + Y = min(X,Y) + max(X,Y),
$$
which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).
answered Jan 18 at 22:50
grand_chatgrand_chat
20.5k11327
$endgroup$
add a comment |
$begingroup$
If you need to prove that
$$
E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
$$
simply note that the relation is valid even before you take expectations:
$$
max(X,Y) = X + Y - min(X,Y),
$$
which is the same as
$$
X + Y = min(X,Y) + max(X,Y),
$$
which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).
answered Jan 18 at 22:50
grand_chatgrand_chat
20.5k11327
$endgroup$
add a comment |
$begingroup$
If you need to prove that
$$
E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
$$
simply note that the relation is valid even before you take expectations:
$$
max(X,Y) = X + Y - min(X,Y),
$$
which is the same as
$$
X + Y = min(X,Y) + max(X,Y),
$$
which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).
answered Jan 18 at 22:50
grand_chatgrand_chat
20.5k11327
$endgroup$
If you need to prove that
$$
E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
$$
simply note that the relation is valid even before you take expectations:
$$
max(X,Y) = X + Y - min(X,Y),
$$
which is the same as
$$
X + Y = min(X,Y) + max(X,Y),
$$
which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).
answered Jan 18 at 22:50
grand_chatgrand_chat
20.5k11327
answered Jan 18 at 22:50
grand_chatgrand_chat
20.5k11327
answered Jan 18 at 22:50
grand_chatgrand_chat
20.5k11327
answered Jan 18 at 22:50
grand_chatgrand_chat
20.5k11327
20.5k11327
add a comment |
add a comment |
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1
$begingroup$
you should porably explain your notations.
$endgroup$
– J.F
Jan 18 at 20:33
$begingroup$
@G.F Iedited the question. Thanks.
$endgroup$
– Laura
Jan 18 at 20:36
1
$begingroup$
I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
$endgroup$
– WaveX
Jan 18 at 20:37
1
$begingroup$
It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
$endgroup$
– JMoravitz
Jan 18 at 20:37
$begingroup$
@JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
$endgroup$
– Laura
Jan 18 at 20:41