Detemine the minimum of two random variable
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The temperatures in Chicago and Detroit are $x^0$ and $y^0$, respectively. These temperatures are not assumed to be independent; namely, we are given: 1) $P(x^0=70)$, 2) $P(y^0 =70)$ and $P(max(x^0,y^0)=70)$. Determine $P(min(x^0,y^0)=70)$.
probability
asked Nov 14 at 23:44
mathnoob
73211
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up vote
1
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The temperatures in Chicago and Detroit are $x^0$ and $y^0$, respectively. These temperatures are not assumed to be independent; namely, we are given: 1) $P(x^0=70)$, 2) $P(y^0 =70)$ and $P(max(x^0,y^0)=70)$. Determine $P(min(x^0,y^0)=70)$.
probability
asked Nov 14 at 23:44
mathnoob
73211
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The temperatures in Chicago and Detroit are $x^0$ and $y^0$, respectively. These temperatures are not assumed to be independent; namely, we are given: 1) $P(x^0=70)$, 2) $P(y^0 =70)$ and $P(max(x^0,y^0)=70)$. Determine $P(min(x^0,y^0)=70)$.
probability
asked Nov 14 at 23:44
mathnoob
73211
The temperatures in Chicago and Detroit are $x^0$ and $y^0$, respectively. These temperatures are not assumed to be independent; namely, we are given: 1) $P(x^0=70)$, 2) $P(y^0 =70)$ and $P(max(x^0,y^0)=70)$. Determine $P(min(x^0,y^0)=70)$.
probability
probability
asked Nov 14 at 23:44
mathnoob
73211
asked Nov 14 at 23:44
mathnoob
73211
asked Nov 14 at 23:44
mathnoob
73211
asked Nov 14 at 23:44
mathnoob
73211
asked Nov 14 at 23:44
mathnoob
73211
73211
add a comment |
add a comment |
1 Answer
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Basic approach. Rewrite $P(x = 70)$ as
$$
P(x = 70) = P(x = 70, y < 70) + P(x = 70, y = 70) + P(x = 70, y > 70)
$$
(do you see why this is true?) and rewrite $P(y = 70)$ similarly. In the same way, rewrite $P(max{x, y} = 70)$ as
$$
P(max{x, y} = 70) = P(x = 70, y < 70) + P(x < 70, y = 70) + P(x = 70, y = 70)
$$
(do you see why this is true?) and rewrite $P(min{x, y} = 70)$ similarly.
The question then becomes: Express the expansion of $P(min{x, y} = 70)$ as a linear combination of the expansions of $P(x = 70)$, $P(y = 70)$, and $P(max{x, y} = 70)$. That is, add and subtract copies of those three quantities, in such a way as to obtain $P(min{x, y} = 70)$.
answered Nov 14 at 23:54
Brian Tung
25.6k32553
So is the solution: $P(X=70)+P(Y=70)-P(max{X,Y}=70)$?
– mathnoob
Nov 15 at 23:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Basic approach. Rewrite $P(x = 70)$ as
$$
P(x = 70) = P(x = 70, y < 70) + P(x = 70, y = 70) + P(x = 70, y > 70)
$$
(do you see why this is true?) and rewrite $P(y = 70)$ similarly. In the same way, rewrite $P(max{x, y} = 70)$ as
$$
P(max{x, y} = 70) = P(x = 70, y < 70) + P(x < 70, y = 70) + P(x = 70, y = 70)
$$
(do you see why this is true?) and rewrite $P(min{x, y} = 70)$ similarly.
The question then becomes: Express the expansion of $P(min{x, y} = 70)$ as a linear combination of the expansions of $P(x = 70)$, $P(y = 70)$, and $P(max{x, y} = 70)$. That is, add and subtract copies of those three quantities, in such a way as to obtain $P(min{x, y} = 70)$.
answered Nov 14 at 23:54
Brian Tung
25.6k32553
So is the solution: $P(X=70)+P(Y=70)-P(max{X,Y}=70)$?
– mathnoob
Nov 15 at 23:32
add a comment |
up vote
0
down vote
accepted
Basic approach. Rewrite $P(x = 70)$ as
$$
P(x = 70) = P(x = 70, y < 70) + P(x = 70, y = 70) + P(x = 70, y > 70)
$$
(do you see why this is true?) and rewrite $P(y = 70)$ similarly. In the same way, rewrite $P(max{x, y} = 70)$ as
$$
P(max{x, y} = 70) = P(x = 70, y < 70) + P(x < 70, y = 70) + P(x = 70, y = 70)
$$
(do you see why this is true?) and rewrite $P(min{x, y} = 70)$ similarly.
The question then becomes: Express the expansion of $P(min{x, y} = 70)$ as a linear combination of the expansions of $P(x = 70)$, $P(y = 70)$, and $P(max{x, y} = 70)$. That is, add and subtract copies of those three quantities, in such a way as to obtain $P(min{x, y} = 70)$.
answered Nov 14 at 23:54
Brian Tung
25.6k32553
So is the solution: $P(X=70)+P(Y=70)-P(max{X,Y}=70)$?
– mathnoob
Nov 15 at 23:32
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Basic approach. Rewrite $P(x = 70)$ as
$$
P(x = 70) = P(x = 70, y < 70) + P(x = 70, y = 70) + P(x = 70, y > 70)
$$
(do you see why this is true?) and rewrite $P(y = 70)$ similarly. In the same way, rewrite $P(max{x, y} = 70)$ as
$$
P(max{x, y} = 70) = P(x = 70, y < 70) + P(x < 70, y = 70) + P(x = 70, y = 70)
$$
(do you see why this is true?) and rewrite $P(min{x, y} = 70)$ similarly.
The question then becomes: Express the expansion of $P(min{x, y} = 70)$ as a linear combination of the expansions of $P(x = 70)$, $P(y = 70)$, and $P(max{x, y} = 70)$. That is, add and subtract copies of those three quantities, in such a way as to obtain $P(min{x, y} = 70)$.
answered Nov 14 at 23:54
Brian Tung
25.6k32553
Basic approach. Rewrite $P(x = 70)$ as
$$
P(x = 70) = P(x = 70, y < 70) + P(x = 70, y = 70) + P(x = 70, y > 70)
$$
(do you see why this is true?) and rewrite $P(y = 70)$ similarly. In the same way, rewrite $P(max{x, y} = 70)$ as
$$
P(max{x, y} = 70) = P(x = 70, y < 70) + P(x < 70, y = 70) + P(x = 70, y = 70)
$$
(do you see why this is true?) and rewrite $P(min{x, y} = 70)$ similarly.
The question then becomes: Express the expansion of $P(min{x, y} = 70)$ as a linear combination of the expansions of $P(x = 70)$, $P(y = 70)$, and $P(max{x, y} = 70)$. That is, add and subtract copies of those three quantities, in such a way as to obtain $P(min{x, y} = 70)$.
answered Nov 14 at 23:54
Brian Tung
25.6k32553
answered Nov 14 at 23:54
Brian Tung
25.6k32553
answered Nov 14 at 23:54
Brian Tung
25.6k32553
answered Nov 14 at 23:54
Brian Tung
25.6k32553
25.6k32553
So is the solution: $P(X=70)+P(Y=70)-P(max{X,Y}=70)$?
– mathnoob
Nov 15 at 23:32
add a comment |
So is the solution: $P(X=70)+P(Y=70)-P(max{X,Y}=70)$?
– mathnoob
Nov 15 at 23:32
So is the solution: $P(X=70)+P(Y=70)-P(max{X,Y}=70)$?
– mathnoob
Nov 15 at 23:32
So is the solution: $P(X=70)+P(Y=70)-P(max{X,Y}=70)$?
– mathnoob
Nov 15 at 23:32
add a comment |
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