Computing marginal distribution
$begingroup$
Assume $x$ is distributed with $F(x)=x^2$ for $0leq x leq 1$ and $cmid x sim U[0, lambda cdot x + 1-lambda]$ for some $0 < lambda < 1$. I am trying to find the marginal distribution of c.
The joint density function should be given by $frac{2x}{lambda cdot x + 1-lambda}$ whenever $x in [0,1]$ and $c in [0, lambda cdot x + 1- lambda]$, right?
I tried integrating out $x$ next, but I am unsure about the boundaries. Simply letting $x geq frac{c+lambda -1}{lambda}$ seems not to be correct, as this lower bound could be negative?
I think should be getting a marginal density for $c$ that integrates to 1 on the support $[0,1]$, regardless of what value $lambda$ takes, but it never worked no matter what I tried...
real-analysis probability integration probability-theory density-function
asked Jan 11 at 3:39
user509037user509037
836
$endgroup$
add a comment |
$begingroup$
Assume $x$ is distributed with $F(x)=x^2$ for $0leq x leq 1$ and $cmid x sim U[0, lambda cdot x + 1-lambda]$ for some $0 < lambda < 1$. I am trying to find the marginal distribution of c.
The joint density function should be given by $frac{2x}{lambda cdot x + 1-lambda}$ whenever $x in [0,1]$ and $c in [0, lambda cdot x + 1- lambda]$, right?
I tried integrating out $x$ next, but I am unsure about the boundaries. Simply letting $x geq frac{c+lambda -1}{lambda}$ seems not to be correct, as this lower bound could be negative?
I think should be getting a marginal density for $c$ that integrates to 1 on the support $[0,1]$, regardless of what value $lambda$ takes, but it never worked no matter what I tried...
real-analysis probability integration probability-theory density-function
asked Jan 11 at 3:39
user509037user509037
836
$endgroup$
$begingroup$
Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
$endgroup$
– JimB
Jan 11 at 7:03
$begingroup$
Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
$endgroup$
– JimB
Jan 11 at 17:07
add a comment |
$begingroup$
Assume $x$ is distributed with $F(x)=x^2$ for $0leq x leq 1$ and $cmid x sim U[0, lambda cdot x + 1-lambda]$ for some $0 < lambda < 1$. I am trying to find the marginal distribution of c.
The joint density function should be given by $frac{2x}{lambda cdot x + 1-lambda}$ whenever $x in [0,1]$ and $c in [0, lambda cdot x + 1- lambda]$, right?
I tried integrating out $x$ next, but I am unsure about the boundaries. Simply letting $x geq frac{c+lambda -1}{lambda}$ seems not to be correct, as this lower bound could be negative?
I think should be getting a marginal density for $c$ that integrates to 1 on the support $[0,1]$, regardless of what value $lambda$ takes, but it never worked no matter what I tried...
real-analysis probability integration probability-theory density-function
asked Jan 11 at 3:39
user509037user509037
836
$endgroup$
Assume $x$ is distributed with $F(x)=x^2$ for $0leq x leq 1$ and $cmid x sim U[0, lambda cdot x + 1-lambda]$ for some $0 < lambda < 1$. I am trying to find the marginal distribution of c.
The joint density function should be given by $frac{2x}{lambda cdot x + 1-lambda}$ whenever $x in [0,1]$ and $c in [0, lambda cdot x + 1- lambda]$, right?
I tried integrating out $x$ next, but I am unsure about the boundaries. Simply letting $x geq frac{c+lambda -1}{lambda}$ seems not to be correct, as this lower bound could be negative?
I think should be getting a marginal density for $c$ that integrates to 1 on the support $[0,1]$, regardless of what value $lambda$ takes, but it never worked no matter what I tried...
real-analysis probability integration probability-theory density-function
real-analysis probability integration probability-theory density-function
asked Jan 11 at 3:39
user509037user509037
836
asked Jan 11 at 3:39
user509037user509037
836
asked Jan 11 at 3:39
user509037user509037
836
asked Jan 11 at 3:39
user509037user509037
836
asked Jan 11 at 3:39
user509037user509037
836
836
$begingroup$
Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
$endgroup$
– JimB
Jan 11 at 7:03
$begingroup$
Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
$endgroup$
– JimB
Jan 11 at 17:07
add a comment |
$begingroup$
Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
$endgroup$
– JimB
Jan 11 at 7:03
$begingroup$
Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
$endgroup$
– JimB
Jan 11 at 17:07
$begingroup$
Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
$endgroup$
– JimB
Jan 11 at 7:03
$begingroup$
Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
$endgroup$
– JimB
Jan 11 at 7:03
$begingroup$
Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
$endgroup$
– JimB
Jan 11 at 17:07
$begingroup$
Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
$endgroup$
– JimB
Jan 11 at 17:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First look at the sample space associated with a non-zero joint density:
The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.
We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.
$$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$
So
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
end{array}
\
end{array}$$
Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:
$$f(c)=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
-frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
end{array}
\
end{array}$$
Here is what the marginal pdf for $C$ looks like for various values of $lambda$:
answered Jan 12 at 5:19
JimBJimB
61547
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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oldest
votes
$begingroup$
First look at the sample space associated with a non-zero joint density:
The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.
We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.
$$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$
So
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
end{array}
\
end{array}$$
Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:
$$f(c)=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
-frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
end{array}
\
end{array}$$
Here is what the marginal pdf for $C$ looks like for various values of $lambda$:
answered Jan 12 at 5:19
JimBJimB
61547
$endgroup$
add a comment |
$begingroup$
First look at the sample space associated with a non-zero joint density:
The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.
We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.
$$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$
So
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
end{array}
\
end{array}$$
Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:
$$f(c)=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
-frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
end{array}
\
end{array}$$
Here is what the marginal pdf for $C$ looks like for various values of $lambda$:
answered Jan 12 at 5:19
JimBJimB
61547
$endgroup$
add a comment |
$begingroup$
First look at the sample space associated with a non-zero joint density:
The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.
We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.
$$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$
So
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
end{array}
\
end{array}$$
Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:
$$f(c)=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
-frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
end{array}
\
end{array}$$
Here is what the marginal pdf for $C$ looks like for various values of $lambda$:
answered Jan 12 at 5:19
JimBJimB
61547
$endgroup$
First look at the sample space associated with a non-zero joint density:
The $1-lambda$ location on the figure is for $lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.
We see that to construct the value of $Pr[Cleq c_0]$ we need to consider two cases: (1) $c_0 <= 1-lambda$ and $c_0 > 1-lambda$.
$$Pr[Cleq c_0 | 0 leq c_0leq 1-lambda leq 1]=int _0^1int _0^{c_0}frac{2 x}{1-lambda (1-x)} dc dx = frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2}$$
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=\ int _0^{frac{c_0+lambda -1}{lambda }}int _0^{1-lambda (1-x)}frac{2 x}{1-lambda (1-x)}dcdx+int _{frac{c_0+lambda -1}{lambda }}^1int _0^{c_0}frac{2 x}{1-lambda (1-x)}dcdx =\
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2}$$
So
$$Pr[Cleq c_0 | 0 lt 1-lambda < c_0 leq 1]=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 c_0 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0leq 1-lambda <1 \
frac{(c_0+lambda -1)^2-2 c_0 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda <c_0<1 \
end{array}
\
end{array}$$
Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:
$$f(c)=begin{array}{cc}
{ &
begin{array}{cc}
frac{2 (lambda -(lambda -1) log (1-lambda ))}{lambda ^2} & 0leq c_0 leq 1-lambda \
-frac{2 ((lambda -1) log (c_0)+c_0-1)}{lambda ^2} & 0<1-lambda< c_0<1 \
end{array}
\
end{array}$$
Here is what the marginal pdf for $C$ looks like for various values of $lambda$:
answered Jan 12 at 5:19
JimBJimB
61547
edited Jan 12 at 5:53
answered Jan 12 at 5:19
JimBJimB
61547
answered Jan 12 at 5:19
JimBJimB
61547
answered Jan 12 at 5:19
JimBJimB
61547
61547
add a comment |
add a comment |
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$begingroup$
Sometimes it is helpful to take a random sample from the joint distribution (randomly selecting an x followed by a value of c conditional on x) and plot the (x,c) pairs along with a histogram of the resulting c values.
$endgroup$
– JimB
Jan 11 at 7:03
$begingroup$
Another hint: Look at the joint distribution and first find $Pr(C leq c)$ as a piecewise function which depends on $c$ being above or below $1-lambda$. Then differentiate the two pieces to get the marginal pdf for $C$.
$endgroup$
– JimB
Jan 11 at 17:07