Joint Density function Question
$begingroup$
The joint probability density function is given as$$f(x,y,z) = kxyz^2$$ where$$ 0<x<1,0<y<1,0<z<2$$
The question asks to find $$P(Z>X+Y)$$
I know we’ll have to find the value of $k$ and I have done that part, but I don’t know how to go forward from there, any help?
probability probability-distributions random-variables
asked Dec 1 '18 at 20:46
user601297user601297
1556
$endgroup$
add a comment |
$begingroup$
The joint probability density function is given as$$f(x,y,z) = kxyz^2$$ where$$ 0<x<1,0<y<1,0<z<2$$
The question asks to find $$P(Z>X+Y)$$
I know we’ll have to find the value of $k$ and I have done that part, but I don’t know how to go forward from there, any help?
probability probability-distributions random-variables
asked Dec 1 '18 at 20:46
user601297user601297
1556
$endgroup$
add a comment |
$begingroup$
The joint probability density function is given as$$f(x,y,z) = kxyz^2$$ where$$ 0<x<1,0<y<1,0<z<2$$
The question asks to find $$P(Z>X+Y)$$
I know we’ll have to find the value of $k$ and I have done that part, but I don’t know how to go forward from there, any help?
probability probability-distributions random-variables
asked Dec 1 '18 at 20:46
user601297user601297
1556
$endgroup$
The joint probability density function is given as$$f(x,y,z) = kxyz^2$$ where$$ 0<x<1,0<y<1,0<z<2$$
The question asks to find $$P(Z>X+Y)$$
I know we’ll have to find the value of $k$ and I have done that part, but I don’t know how to go forward from there, any help?
probability probability-distributions random-variables
probability probability-distributions random-variables
asked Dec 1 '18 at 20:46
user601297user601297
1556
asked Dec 1 '18 at 20:46
user601297user601297
1556
asked Dec 1 '18 at 20:46
user601297user601297
1556
asked Dec 1 '18 at 20:46
user601297user601297
1556
asked Dec 1 '18 at 20:46
user601297user601297
1556
1556
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You just have to integrate the density over the region where $z>x+y$ so$$int_0^1int_0^1int_{x+y}^2f(x,y,z)dzdydx$$
answered Dec 1 '18 at 20:59
saulspatzsaulspatz
14.2k21329
$endgroup$
$begingroup$
Thanks i got this.
$endgroup$
– user601297
Dec 1 '18 at 21:21
add a comment |
$begingroup$
As this seems like a homework question I won't answer your question completely, but I will give you enough help so that you could figure it out from here. Please ask for further clarification if anything is unclear.
I assume $X$, $Y$ and $Z$ are independent. You will need the marginal distributions of all three variables. Since you managed to determine $k$ I assume you know how to get these. Then consider the following assertion:
You don't care about any particular values of $X$ or $Y$. You simply care that, given any two values, what is the probability that $Z$ is larger? Say $X = x$ and $Y = y$. Then this probability will be $int_{x+y}^{2} f_Z(z) mathop{mathrm{d}z}$. If this does not make sense to you, you should take the time to figure it out.
However, you need to account for the fact that $x$ and $y$ are unknown. You can do this by integrating over every possible $x$ and $y$. This also allows you to take into account that not every $x$ is equally probable. This means you will end up having to evaluate a triple integral.
answered Dec 1 '18 at 21:03
Erik AndréErik André
857
$endgroup$
$begingroup$
Yes I did exactly this, but won’t the final answer in this case be in terms of x and y? If yes is that correct, that is reall my only condusion.
$endgroup$
– user601297
Dec 1 '18 at 21:21
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
$begingroup$
You just have to integrate the density over the region where $z>x+y$ so$$int_0^1int_0^1int_{x+y}^2f(x,y,z)dzdydx$$
answered Dec 1 '18 at 20:59
saulspatzsaulspatz
14.2k21329
$endgroup$
$begingroup$
Thanks i got this.
$endgroup$
– user601297
Dec 1 '18 at 21:21
add a comment |
$begingroup$
You just have to integrate the density over the region where $z>x+y$ so$$int_0^1int_0^1int_{x+y}^2f(x,y,z)dzdydx$$
answered Dec 1 '18 at 20:59
saulspatzsaulspatz
14.2k21329
$endgroup$
$begingroup$
Thanks i got this.
$endgroup$
– user601297
Dec 1 '18 at 21:21
add a comment |
$begingroup$
You just have to integrate the density over the region where $z>x+y$ so$$int_0^1int_0^1int_{x+y}^2f(x,y,z)dzdydx$$
answered Dec 1 '18 at 20:59
saulspatzsaulspatz
14.2k21329
$endgroup$
You just have to integrate the density over the region where $z>x+y$ so$$int_0^1int_0^1int_{x+y}^2f(x,y,z)dzdydx$$
answered Dec 1 '18 at 20:59
saulspatzsaulspatz
14.2k21329
answered Dec 1 '18 at 20:59
saulspatzsaulspatz
14.2k21329
answered Dec 1 '18 at 20:59
saulspatzsaulspatz
14.2k21329
answered Dec 1 '18 at 20:59
saulspatzsaulspatz
14.2k21329
14.2k21329
$begingroup$
Thanks i got this.
$endgroup$
– user601297
Dec 1 '18 at 21:21
add a comment |
$begingroup$
Thanks i got this.
$endgroup$
– user601297
Dec 1 '18 at 21:21
$begingroup$
Thanks i got this.
$endgroup$
– user601297
Dec 1 '18 at 21:21
$begingroup$
Thanks i got this.
$endgroup$
– user601297
Dec 1 '18 at 21:21
add a comment |
$begingroup$
As this seems like a homework question I won't answer your question completely, but I will give you enough help so that you could figure it out from here. Please ask for further clarification if anything is unclear.
I assume $X$, $Y$ and $Z$ are independent. You will need the marginal distributions of all three variables. Since you managed to determine $k$ I assume you know how to get these. Then consider the following assertion:
You don't care about any particular values of $X$ or $Y$. You simply care that, given any two values, what is the probability that $Z$ is larger? Say $X = x$ and $Y = y$. Then this probability will be $int_{x+y}^{2} f_Z(z) mathop{mathrm{d}z}$. If this does not make sense to you, you should take the time to figure it out.
However, you need to account for the fact that $x$ and $y$ are unknown. You can do this by integrating over every possible $x$ and $y$. This also allows you to take into account that not every $x$ is equally probable. This means you will end up having to evaluate a triple integral.
answered Dec 1 '18 at 21:03
Erik AndréErik André
857
$endgroup$
$begingroup$
Yes I did exactly this, but won’t the final answer in this case be in terms of x and y? If yes is that correct, that is reall my only condusion.
$endgroup$
– user601297
Dec 1 '18 at 21:21
add a comment |
$begingroup$
As this seems like a homework question I won't answer your question completely, but I will give you enough help so that you could figure it out from here. Please ask for further clarification if anything is unclear.
I assume $X$, $Y$ and $Z$ are independent. You will need the marginal distributions of all three variables. Since you managed to determine $k$ I assume you know how to get these. Then consider the following assertion:
You don't care about any particular values of $X$ or $Y$. You simply care that, given any two values, what is the probability that $Z$ is larger? Say $X = x$ and $Y = y$. Then this probability will be $int_{x+y}^{2} f_Z(z) mathop{mathrm{d}z}$. If this does not make sense to you, you should take the time to figure it out.
However, you need to account for the fact that $x$ and $y$ are unknown. You can do this by integrating over every possible $x$ and $y$. This also allows you to take into account that not every $x$ is equally probable. This means you will end up having to evaluate a triple integral.
answered Dec 1 '18 at 21:03
Erik AndréErik André
857
$endgroup$
$begingroup$
Yes I did exactly this, but won’t the final answer in this case be in terms of x and y? If yes is that correct, that is reall my only condusion.
$endgroup$
– user601297
Dec 1 '18 at 21:21
add a comment |
$begingroup$
As this seems like a homework question I won't answer your question completely, but I will give you enough help so that you could figure it out from here. Please ask for further clarification if anything is unclear.
I assume $X$, $Y$ and $Z$ are independent. You will need the marginal distributions of all three variables. Since you managed to determine $k$ I assume you know how to get these. Then consider the following assertion:
You don't care about any particular values of $X$ or $Y$. You simply care that, given any two values, what is the probability that $Z$ is larger? Say $X = x$ and $Y = y$. Then this probability will be $int_{x+y}^{2} f_Z(z) mathop{mathrm{d}z}$. If this does not make sense to you, you should take the time to figure it out.
However, you need to account for the fact that $x$ and $y$ are unknown. You can do this by integrating over every possible $x$ and $y$. This also allows you to take into account that not every $x$ is equally probable. This means you will end up having to evaluate a triple integral.
answered Dec 1 '18 at 21:03
Erik AndréErik André
857
$endgroup$
As this seems like a homework question I won't answer your question completely, but I will give you enough help so that you could figure it out from here. Please ask for further clarification if anything is unclear.
I assume $X$, $Y$ and $Z$ are independent. You will need the marginal distributions of all three variables. Since you managed to determine $k$ I assume you know how to get these. Then consider the following assertion:
You don't care about any particular values of $X$ or $Y$. You simply care that, given any two values, what is the probability that $Z$ is larger? Say $X = x$ and $Y = y$. Then this probability will be $int_{x+y}^{2} f_Z(z) mathop{mathrm{d}z}$. If this does not make sense to you, you should take the time to figure it out.
However, you need to account for the fact that $x$ and $y$ are unknown. You can do this by integrating over every possible $x$ and $y$. This also allows you to take into account that not every $x$ is equally probable. This means you will end up having to evaluate a triple integral.
answered Dec 1 '18 at 21:03
Erik AndréErik André
857
answered Dec 1 '18 at 21:03
Erik AndréErik André
857
answered Dec 1 '18 at 21:03
Erik AndréErik André
857
answered Dec 1 '18 at 21:03
Erik AndréErik André
857
857
$begingroup$
Yes I did exactly this, but won’t the final answer in this case be in terms of x and y? If yes is that correct, that is reall my only condusion.
$endgroup$
– user601297
Dec 1 '18 at 21:21
add a comment |
$begingroup$
Yes I did exactly this, but won’t the final answer in this case be in terms of x and y? If yes is that correct, that is reall my only condusion.
$endgroup$
– user601297
Dec 1 '18 at 21:21
$begingroup$
Yes I did exactly this, but won’t the final answer in this case be in terms of x and y? If yes is that correct, that is reall my only condusion.
$endgroup$
– user601297
Dec 1 '18 at 21:21
$begingroup$
Yes I did exactly this, but won’t the final answer in this case be in terms of x and y? If yes is that correct, that is reall my only condusion.
$endgroup$
– user601297
Dec 1 '18 at 21:21
add a comment |
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