Mixing
Distribution of $X_1X_2X_3$ if $(X_1,X_2,X_3)$ has PDF $frac{2}{2e-5}x_1^2x_2e^{x_1x_2x_3}$ on $0<...
$begingroup$
The random vector X = $(X_1,X_2,X_3)'$ has density function
$f_{X}(x)=begin{cases}
frac{2}{2e-5}.x_1^2.x_2.e^{x_1.x_2.x_3}, & for & 0 < x_1,x_2,x_3<1 \[2ex]
0, & text{otherwise}
end{cases}$
Determine the distribution of $X_1.X_2.X_3$:
I've tried getting the marginals and multiply them to get the pdf of
$Y=X_1.X_2.X_3$
but I'm not sure if that could be done since I'm not sure if I can assume that $X_i$'s are all independent.
My main issue is with the bounds with min. I can't figure those out.
probability-theory probability-distributions
asked Jan 27 at 10:40
Mahamad A. KanoutéMahamad A. Kanouté
399114
$endgroup$
add a comment |
$begingroup$
The random vector X = $(X_1,X_2,X_3)'$ has density function
$f_{X}(x)=begin{cases}
frac{2}{2e-5}.x_1^2.x_2.e^{x_1.x_2.x_3}, & for & 0 < x_1,x_2,x_3<1 \[2ex]
0, & text{otherwise}
end{cases}$
Determine the distribution of $X_1.X_2.X_3$:
I've tried getting the marginals and multiply them to get the pdf of
$Y=X_1.X_2.X_3$
but I'm not sure if that could be done since I'm not sure if I can assume that $X_i$'s are all independent.
My main issue is with the bounds with min. I can't figure those out.
probability-theory probability-distributions
asked Jan 27 at 10:40
Mahamad A. KanoutéMahamad A. Kanouté
399114
$endgroup$
1
$begingroup$
They are not independent. Even if they were, multiplying the marginal densities would not give you the density of the product
$endgroup$
– Henry
Jan 27 at 10:45
add a comment |
$begingroup$
The random vector X = $(X_1,X_2,X_3)'$ has density function
$f_{X}(x)=begin{cases}
frac{2}{2e-5}.x_1^2.x_2.e^{x_1.x_2.x_3}, & for & 0 < x_1,x_2,x_3<1 \[2ex]
0, & text{otherwise}
end{cases}$
Determine the distribution of $X_1.X_2.X_3$:
I've tried getting the marginals and multiply them to get the pdf of
$Y=X_1.X_2.X_3$
but I'm not sure if that could be done since I'm not sure if I can assume that $X_i$'s are all independent.
My main issue is with the bounds with min. I can't figure those out.
probability-theory probability-distributions
asked Jan 27 at 10:40
Mahamad A. KanoutéMahamad A. Kanouté
399114
$endgroup$
The random vector X = $(X_1,X_2,X_3)'$ has density function
$f_{X}(x)=begin{cases}
frac{2}{2e-5}.x_1^2.x_2.e^{x_1.x_2.x_3}, & for & 0 < x_1,x_2,x_3<1 \[2ex]
0, & text{otherwise}
end{cases}$
Determine the distribution of $X_1.X_2.X_3$:
I've tried getting the marginals and multiply them to get the pdf of
$Y=X_1.X_2.X_3$
but I'm not sure if that could be done since I'm not sure if I can assume that $X_i$'s are all independent.
My main issue is with the bounds with min. I can't figure those out.
probability-theory probability-distributions
probability-theory probability-distributions
asked Jan 27 at 10:40
Mahamad A. KanoutéMahamad A. Kanouté
399114
asked Jan 27 at 10:40
Mahamad A. KanoutéMahamad A. Kanouté
399114
edited Jan 28 at 3:13
Mahamad A. Kanouté
asked Jan 27 at 10:40
Mahamad A. KanoutéMahamad A. Kanouté
399114
asked Jan 27 at 10:40
Mahamad A. KanoutéMahamad A. Kanouté
399114
asked Jan 27 at 10:40
Mahamad A. KanoutéMahamad A. Kanouté
399114
399114
1
$begingroup$
They are not independent. Even if they were, multiplying the marginal densities would not give you the density of the product
$endgroup$
– Henry
Jan 27 at 10:45
add a comment |
1
$begingroup$
They are not independent. Even if they were, multiplying the marginal densities would not give you the density of the product
$endgroup$
– Henry
Jan 27 at 10:45
1
1
$begingroup$
They are not independent. Even if they were, multiplying the marginal densities would not give you the density of the product
$endgroup$
– Henry
Jan 27 at 10:45
$begingroup$
They are not independent. Even if they were, multiplying the marginal densities would not give you the density of the product
$endgroup$
– Henry
Jan 27 at 10:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First we write the CDF of $Y=X_1X_2X_3$, $F_Y(y)=P(Yleq y)$ when $0leq y leq 1$:
$$F_Y(y)=int_0^{1}int_0^1int_0^1f(x_1,x_2,x_3)mathbf{I}(x_1x_2x_3leq y)dx_1dx_2dx_3$$
This can be written in three regions: (simplify the notation as $f(mathbf{x})$ and $dmathbf{x}$ to denote $f(x_1,x_2,x_3)$ and $dx_1dx_2dx_3$, in the same order.
$$F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$$
Performing the integrations, if I don't have any computational mistakes, this CDF can be found as
$$F_Y(y)=frac{e^y(y^2-4y+5)-5}{2e-5}$$
which yields (for $0leq y leq 1$)
$$f_Y(y)=frac{dF_Y(y)}{dy}=frac{e^y(y-1)^2}{2e-5}$$
answered Jan 27 at 12:29
gunesgunes
3747
$endgroup$
$begingroup$
my main issue is with the bounds that you have with min. I can't figure those out
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:05
$begingroup$
like going from the first line to the second
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:06
$begingroup$
I actually went from second to first :) while trying to represent it more compact, introduced a mistake. Now edited. However, the second integral is correct. Do you have problems with it?
$endgroup$
– gunes
Jan 28 at 4:28
$begingroup$
I now have a problem with the $F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 4:38
1
$begingroup$
The first integral: When $x_1<y$, $x_2,x_3$ can be anything in $[0,1]$, still $x_1x_2x_3<y$. When $x_1>y$, there are two options. Second integral: If $x_2<y/x_1$, $x_3$ can be anything since $x_1x_2<y$. Third integral: When $x_2>y/x_1$, $x_1x_2$ becomes $>y$, and we need to consider $x_3$ 's where $x_1x_2x_3<y$. To be able to do that, we need to choose $x_3<y/x_1x_2$.
$endgroup$
– gunes
Jan 28 at 4:44
|
show 2 more comments
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
First we write the CDF of $Y=X_1X_2X_3$, $F_Y(y)=P(Yleq y)$ when $0leq y leq 1$:
$$F_Y(y)=int_0^{1}int_0^1int_0^1f(x_1,x_2,x_3)mathbf{I}(x_1x_2x_3leq y)dx_1dx_2dx_3$$
This can be written in three regions: (simplify the notation as $f(mathbf{x})$ and $dmathbf{x}$ to denote $f(x_1,x_2,x_3)$ and $dx_1dx_2dx_3$, in the same order.
$$F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$$
Performing the integrations, if I don't have any computational mistakes, this CDF can be found as
$$F_Y(y)=frac{e^y(y^2-4y+5)-5}{2e-5}$$
which yields (for $0leq y leq 1$)
$$f_Y(y)=frac{dF_Y(y)}{dy}=frac{e^y(y-1)^2}{2e-5}$$
answered Jan 27 at 12:29
gunesgunes
3747
$endgroup$
$begingroup$
my main issue is with the bounds that you have with min. I can't figure those out
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:05
$begingroup$
like going from the first line to the second
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:06
$begingroup$
I actually went from second to first :) while trying to represent it more compact, introduced a mistake. Now edited. However, the second integral is correct. Do you have problems with it?
$endgroup$
– gunes
Jan 28 at 4:28
$begingroup$
I now have a problem with the $F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 4:38
1
$begingroup$
The first integral: When $x_1<y$, $x_2,x_3$ can be anything in $[0,1]$, still $x_1x_2x_3<y$. When $x_1>y$, there are two options. Second integral: If $x_2<y/x_1$, $x_3$ can be anything since $x_1x_2<y$. Third integral: When $x_2>y/x_1$, $x_1x_2$ becomes $>y$, and we need to consider $x_3$ 's where $x_1x_2x_3<y$. To be able to do that, we need to choose $x_3<y/x_1x_2$.
$endgroup$
– gunes
Jan 28 at 4:44
|
show 2 more comments
$begingroup$
First we write the CDF of $Y=X_1X_2X_3$, $F_Y(y)=P(Yleq y)$ when $0leq y leq 1$:
$$F_Y(y)=int_0^{1}int_0^1int_0^1f(x_1,x_2,x_3)mathbf{I}(x_1x_2x_3leq y)dx_1dx_2dx_3$$
This can be written in three regions: (simplify the notation as $f(mathbf{x})$ and $dmathbf{x}$ to denote $f(x_1,x_2,x_3)$ and $dx_1dx_2dx_3$, in the same order.
$$F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$$
Performing the integrations, if I don't have any computational mistakes, this CDF can be found as
$$F_Y(y)=frac{e^y(y^2-4y+5)-5}{2e-5}$$
which yields (for $0leq y leq 1$)
$$f_Y(y)=frac{dF_Y(y)}{dy}=frac{e^y(y-1)^2}{2e-5}$$
answered Jan 27 at 12:29
gunesgunes
3747
$endgroup$
$begingroup$
my main issue is with the bounds that you have with min. I can't figure those out
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:05
$begingroup$
like going from the first line to the second
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:06
$begingroup$
I actually went from second to first :) while trying to represent it more compact, introduced a mistake. Now edited. However, the second integral is correct. Do you have problems with it?
$endgroup$
– gunes
Jan 28 at 4:28
$begingroup$
I now have a problem with the $F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 4:38
1
$begingroup$
The first integral: When $x_1<y$, $x_2,x_3$ can be anything in $[0,1]$, still $x_1x_2x_3<y$. When $x_1>y$, there are two options. Second integral: If $x_2<y/x_1$, $x_3$ can be anything since $x_1x_2<y$. Third integral: When $x_2>y/x_1$, $x_1x_2$ becomes $>y$, and we need to consider $x_3$ 's where $x_1x_2x_3<y$. To be able to do that, we need to choose $x_3<y/x_1x_2$.
$endgroup$
– gunes
Jan 28 at 4:44
|
show 2 more comments
$begingroup$
First we write the CDF of $Y=X_1X_2X_3$, $F_Y(y)=P(Yleq y)$ when $0leq y leq 1$:
$$F_Y(y)=int_0^{1}int_0^1int_0^1f(x_1,x_2,x_3)mathbf{I}(x_1x_2x_3leq y)dx_1dx_2dx_3$$
This can be written in three regions: (simplify the notation as $f(mathbf{x})$ and $dmathbf{x}$ to denote $f(x_1,x_2,x_3)$ and $dx_1dx_2dx_3$, in the same order.
$$F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$$
Performing the integrations, if I don't have any computational mistakes, this CDF can be found as
$$F_Y(y)=frac{e^y(y^2-4y+5)-5}{2e-5}$$
which yields (for $0leq y leq 1$)
$$f_Y(y)=frac{dF_Y(y)}{dy}=frac{e^y(y-1)^2}{2e-5}$$
answered Jan 27 at 12:29
gunesgunes
3747
$endgroup$
First we write the CDF of $Y=X_1X_2X_3$, $F_Y(y)=P(Yleq y)$ when $0leq y leq 1$:
$$F_Y(y)=int_0^{1}int_0^1int_0^1f(x_1,x_2,x_3)mathbf{I}(x_1x_2x_3leq y)dx_1dx_2dx_3$$
This can be written in three regions: (simplify the notation as $f(mathbf{x})$ and $dmathbf{x}$ to denote $f(x_1,x_2,x_3)$ and $dx_1dx_2dx_3$, in the same order.
$$F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$$
Performing the integrations, if I don't have any computational mistakes, this CDF can be found as
$$F_Y(y)=frac{e^y(y^2-4y+5)-5}{2e-5}$$
which yields (for $0leq y leq 1$)
$$f_Y(y)=frac{dF_Y(y)}{dy}=frac{e^y(y-1)^2}{2e-5}$$
answered Jan 27 at 12:29
gunesgunes
3747
edited Jan 28 at 4:45
answered Jan 27 at 12:29
gunesgunes
3747
answered Jan 27 at 12:29
gunesgunes
3747
answered Jan 27 at 12:29
gunesgunes
3747
3747
$begingroup$
my main issue is with the bounds that you have with min. I can't figure those out
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:05
$begingroup$
like going from the first line to the second
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:06
$begingroup$
I actually went from second to first :) while trying to represent it more compact, introduced a mistake. Now edited. However, the second integral is correct. Do you have problems with it?
$endgroup$
– gunes
Jan 28 at 4:28
$begingroup$
I now have a problem with the $F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 4:38
1
$begingroup$
The first integral: When $x_1<y$, $x_2,x_3$ can be anything in $[0,1]$, still $x_1x_2x_3<y$. When $x_1>y$, there are two options. Second integral: If $x_2<y/x_1$, $x_3$ can be anything since $x_1x_2<y$. Third integral: When $x_2>y/x_1$, $x_1x_2$ becomes $>y$, and we need to consider $x_3$ 's where $x_1x_2x_3<y$. To be able to do that, we need to choose $x_3<y/x_1x_2$.
$endgroup$
– gunes
Jan 28 at 4:44
|
show 2 more comments
$begingroup$
my main issue is with the bounds that you have with min. I can't figure those out
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:05
$begingroup$
like going from the first line to the second
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:06
$begingroup$
I actually went from second to first :) while trying to represent it more compact, introduced a mistake. Now edited. However, the second integral is correct. Do you have problems with it?
$endgroup$
– gunes
Jan 28 at 4:28
$begingroup$
I now have a problem with the $F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 4:38
1
$begingroup$
The first integral: When $x_1<y$, $x_2,x_3$ can be anything in $[0,1]$, still $x_1x_2x_3<y$. When $x_1>y$, there are two options. Second integral: If $x_2<y/x_1$, $x_3$ can be anything since $x_1x_2<y$. Third integral: When $x_2>y/x_1$, $x_1x_2$ becomes $>y$, and we need to consider $x_3$ 's where $x_1x_2x_3<y$. To be able to do that, we need to choose $x_3<y/x_1x_2$.
$endgroup$
– gunes
Jan 28 at 4:44
$begingroup$
my main issue is with the bounds that you have with min. I can't figure those out
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:05
$begingroup$
my main issue is with the bounds that you have with min. I can't figure those out
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:05
$begingroup$
like going from the first line to the second
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:06
$begingroup$
like going from the first line to the second
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 3:06
$begingroup$
I actually went from second to first :) while trying to represent it more compact, introduced a mistake. Now edited. However, the second integral is correct. Do you have problems with it?
$endgroup$
– gunes
Jan 28 at 4:28
$begingroup$
I actually went from second to first :) while trying to represent it more compact, introduced a mistake. Now edited. However, the second integral is correct. Do you have problems with it?
$endgroup$
– gunes
Jan 28 at 4:28
$begingroup$
I now have a problem with the $F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 4:38
$begingroup$
I now have a problem with the $F_Y(y)=int_0^yint_0^1int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_0^{y/x_1}int_0^1f(mathbf{x})dmathbf{x}+int_y^1int_{y/x_1}^1int_0^{y/x_1x_2}f(mathbf{x})dmathbf{x}$
$endgroup$
– Mahamad A. Kanouté
Jan 28 at 4:38
1
1
$begingroup$
The first integral: When $x_1<y$, $x_2,x_3$ can be anything in $[0,1]$, still $x_1x_2x_3<y$. When $x_1>y$, there are two options. Second integral: If $x_2<y/x_1$, $x_3$ can be anything since $x_1x_2<y$. Third integral: When $x_2>y/x_1$, $x_1x_2$ becomes $>y$, and we need to consider $x_3$ 's where $x_1x_2x_3<y$. To be able to do that, we need to choose $x_3<y/x_1x_2$.
$endgroup$
– gunes
Jan 28 at 4:44
$begingroup$
The first integral: When $x_1<y$, $x_2,x_3$ can be anything in $[0,1]$, still $x_1x_2x_3<y$. When $x_1>y$, there are two options. Second integral: If $x_2<y/x_1$, $x_3$ can be anything since $x_1x_2<y$. Third integral: When $x_2>y/x_1$, $x_1x_2$ becomes $>y$, and we need to consider $x_3$ 's where $x_1x_2x_3<y$. To be able to do that, we need to choose $x_3<y/x_1x_2$.
$endgroup$
– gunes
Jan 28 at 4:44
|
show 2 more comments
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$begingroup$
They are not independent. Even if they were, multiplying the marginal densities would not give you the density of the product
$endgroup$
– Henry
Jan 27 at 10:45