Mixing
probability from joint density function [closed]
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Could anyone help with this problem? Thanks
A joint density function is given as follows:
$$f(x,y) =begincases 0.125cdot (x+y+1) textfor -1<x<1, 0<y<2 \ 0, textotherwise endcases$$
Calculate $P(X>Y)$
density-function
edited Jul 25 at 9:45
callculus
16.4k31427
asked Jul 25 at 9:36
Denson
367
closed as off-topic by Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly Taxel Jul 26 at 4:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly Taxel
add a comment |Â
up vote
0
down vote
favorite
Could anyone help with this problem? Thanks
A joint density function is given as follows:
$$f(x,y) =begincases 0.125cdot (x+y+1) textfor -1<x<1, 0<y<2 \ 0, textotherwise endcases$$
Calculate $P(X>Y)$
density-function
edited Jul 25 at 9:45
callculus
16.4k31427
asked Jul 25 at 9:36
Denson
367
closed as off-topic by Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly Taxel Jul 26 at 4:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly Taxel
There is something missing here. Where do we define the function according to the formula? Surely not on the whole plane, it would not be a probability density function.
– A. Pongrácz
Jul 25 at 9:39
@A.Pongrácz I´ve fixed it.
– callculus
Jul 25 at 9:40
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Could anyone help with this problem? Thanks
A joint density function is given as follows:
$$f(x,y) =begincases 0.125cdot (x+y+1) textfor -1<x<1, 0<y<2 \ 0, textotherwise endcases$$
Calculate $P(X>Y)$
density-function
edited Jul 25 at 9:45
callculus
16.4k31427
asked Jul 25 at 9:36
Denson
367
Could anyone help with this problem? Thanks
A joint density function is given as follows:
$$f(x,y) =begincases 0.125cdot (x+y+1) textfor -1<x<1, 0<y<2 \ 0, textotherwise endcases$$
Calculate $P(X>Y)$
density-function
edited Jul 25 at 9:45
callculus
16.4k31427
asked Jul 25 at 9:36
Denson
367
edited Jul 25 at 9:45
callculus
16.4k31427
edited Jul 25 at 9:45
callculus
16.4k31427
edited Jul 25 at 9:45
callculus
16.4k31427
16.4k31427
asked Jul 25 at 9:36
Denson
367
asked Jul 25 at 9:36
Denson
367
asked Jul 25 at 9:36
Denson
367
367
closed as off-topic by Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly Taxel Jul 26 at 4:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly Taxel
closed as off-topic by Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly Taxel Jul 26 at 4:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Xander Henderson, Isaac Browne, Nils Matthes, Parcly Taxel
There is something missing here. Where do we define the function according to the formula? Surely not on the whole plane, it would not be a probability density function.
– A. Pongrácz
Jul 25 at 9:39
@A.Pongrácz I´ve fixed it.
– callculus
Jul 25 at 9:40
add a comment |Â
There is something missing here. Where do we define the function according to the formula? Surely not on the whole plane, it would not be a probability density function.
– A. Pongrácz
Jul 25 at 9:39
@A.Pongrácz I´ve fixed it.
– callculus
Jul 25 at 9:40
There is something missing here. Where do we define the function according to the formula? Surely not on the whole plane, it would not be a probability density function.
– A. Pongrácz
Jul 25 at 9:39
There is something missing here. Where do we define the function according to the formula? Surely not on the whole plane, it would not be a probability density function.
– A. Pongrácz
Jul 25 at 9:39
@A.Pongrácz I´ve fixed it.
– callculus
Jul 25 at 9:40
@A.Pongrácz I´ve fixed it.
– callculus
Jul 25 at 9:40
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Just recall what the density function represents: the probability of an event $A$ is the integral of the density function on $A$.
So you have to inegrate the function on the set of points $A= (x,y) mid x>y $. So $x$ can be any number in $[-1, 1]$, and $y$ has to be smaller than $x$.
Hence, compute $intlimits_-1^1 intlimits_0^x f(x,y) , dy , dx$.
As you integrate 0 in the inner integral whenever $x$ is negative, it is the same as $intlimits_0^1 intlimits_0^x f(x,y) , dy , dx$. You can easily compute this.
answered Jul 25 at 9:45
A. Pongrácz
1,804116
Thanks very much. I got stuck with the limits of the integrals- if the limit of the inner integral is from 0 to x, then I thought this would imply that x can be equal to y but we are asked to calculate P(X>Y) (and not P(X>=Y))
– Denson
Jul 25 at 10:05
Formally speaking, you are right. But the densitiy function is continuous, so it doesn't matter if you include/exclude points or even curves in/from the event, it does not affect the result.
– A. Pongrácz
Jul 25 at 11:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Just recall what the density function represents: the probability of an event $A$ is the integral of the density function on $A$.
So you have to inegrate the function on the set of points $A= (x,y) mid x>y $. So $x$ can be any number in $[-1, 1]$, and $y$ has to be smaller than $x$.
Hence, compute $intlimits_-1^1 intlimits_0^x f(x,y) , dy , dx$.
As you integrate 0 in the inner integral whenever $x$ is negative, it is the same as $intlimits_0^1 intlimits_0^x f(x,y) , dy , dx$. You can easily compute this.
answered Jul 25 at 9:45
A. Pongrácz
1,804116
Thanks very much. I got stuck with the limits of the integrals- if the limit of the inner integral is from 0 to x, then I thought this would imply that x can be equal to y but we are asked to calculate P(X>Y) (and not P(X>=Y))
– Denson
Jul 25 at 10:05
Formally speaking, you are right. But the densitiy function is continuous, so it doesn't matter if you include/exclude points or even curves in/from the event, it does not affect the result.
– A. Pongrácz
Jul 25 at 11:39
add a comment |Â
up vote
2
down vote
accepted
Just recall what the density function represents: the probability of an event $A$ is the integral of the density function on $A$.
So you have to inegrate the function on the set of points $A= (x,y) mid x>y $. So $x$ can be any number in $[-1, 1]$, and $y$ has to be smaller than $x$.
Hence, compute $intlimits_-1^1 intlimits_0^x f(x,y) , dy , dx$.
As you integrate 0 in the inner integral whenever $x$ is negative, it is the same as $intlimits_0^1 intlimits_0^x f(x,y) , dy , dx$. You can easily compute this.
answered Jul 25 at 9:45
A. Pongrácz
1,804116
Thanks very much. I got stuck with the limits of the integrals- if the limit of the inner integral is from 0 to x, then I thought this would imply that x can be equal to y but we are asked to calculate P(X>Y) (and not P(X>=Y))
– Denson
Jul 25 at 10:05
Formally speaking, you are right. But the densitiy function is continuous, so it doesn't matter if you include/exclude points or even curves in/from the event, it does not affect the result.
– A. Pongrácz
Jul 25 at 11:39
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Just recall what the density function represents: the probability of an event $A$ is the integral of the density function on $A$.
So you have to inegrate the function on the set of points $A= (x,y) mid x>y $. So $x$ can be any number in $[-1, 1]$, and $y$ has to be smaller than $x$.
Hence, compute $intlimits_-1^1 intlimits_0^x f(x,y) , dy , dx$.
As you integrate 0 in the inner integral whenever $x$ is negative, it is the same as $intlimits_0^1 intlimits_0^x f(x,y) , dy , dx$. You can easily compute this.
answered Jul 25 at 9:45
A. Pongrácz
1,804116
Just recall what the density function represents: the probability of an event $A$ is the integral of the density function on $A$.
So you have to inegrate the function on the set of points $A= (x,y) mid x>y $. So $x$ can be any number in $[-1, 1]$, and $y$ has to be smaller than $x$.
Hence, compute $intlimits_-1^1 intlimits_0^x f(x,y) , dy , dx$.
As you integrate 0 in the inner integral whenever $x$ is negative, it is the same as $intlimits_0^1 intlimits_0^x f(x,y) , dy , dx$. You can easily compute this.
answered Jul 25 at 9:45
A. Pongrácz
1,804116
answered Jul 25 at 9:45
A. Pongrácz
1,804116
answered Jul 25 at 9:45
A. Pongrácz
1,804116
answered Jul 25 at 9:45
A. Pongrácz
1,804116
1,804116
Thanks very much. I got stuck with the limits of the integrals- if the limit of the inner integral is from 0 to x, then I thought this would imply that x can be equal to y but we are asked to calculate P(X>Y) (and not P(X>=Y))
– Denson
Jul 25 at 10:05
Formally speaking, you are right. But the densitiy function is continuous, so it doesn't matter if you include/exclude points or even curves in/from the event, it does not affect the result.
– A. Pongrácz
Jul 25 at 11:39
add a comment |Â
Thanks very much. I got stuck with the limits of the integrals- if the limit of the inner integral is from 0 to x, then I thought this would imply that x can be equal to y but we are asked to calculate P(X>Y) (and not P(X>=Y))
– Denson
Jul 25 at 10:05
Formally speaking, you are right. But the densitiy function is continuous, so it doesn't matter if you include/exclude points or even curves in/from the event, it does not affect the result.
– A. Pongrácz
Jul 25 at 11:39
Thanks very much. I got stuck with the limits of the integrals- if the limit of the inner integral is from 0 to x, then I thought this would imply that x can be equal to y but we are asked to calculate P(X>Y) (and not P(X>=Y))
– Denson
Jul 25 at 10:05
Thanks very much. I got stuck with the limits of the integrals- if the limit of the inner integral is from 0 to x, then I thought this would imply that x can be equal to y but we are asked to calculate P(X>Y) (and not P(X>=Y))
– Denson
Jul 25 at 10:05
Formally speaking, you are right. But the densitiy function is continuous, so it doesn't matter if you include/exclude points or even curves in/from the event, it does not affect the result.
– A. Pongrácz
Jul 25 at 11:39
Formally speaking, you are right. But the densitiy function is continuous, so it doesn't matter if you include/exclude points or even curves in/from the event, it does not affect the result.
– A. Pongrácz
Jul 25 at 11:39
add a comment |Â
There is something missing here. Where do we define the function according to the formula? Surely not on the whole plane, it would not be a probability density function.
– A. Pongrácz
Jul 25 at 9:39
@A.Pongrácz I´ve fixed it.
– callculus
Jul 25 at 9:40