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joint density of two sums of independent random var with common component

1

$begingroup$

Suppose we have three iid draws from a uniform distribution on $[0,1]$. Call these random variables $A, B$ and $C$.
Let $X=A+B$ and $Y=B+C$.
I have figured out that the density of $X$ (or $Y$) is
$$f_X(x) = begincases
x &mbox if x in [0,1] \
2-x &mbox if x in (1,2].
endcases$$
I can also see that $X=A+Y-C$.
However, I am still struggeling with the joint density of $X,Y$ and the conditional density of $X|Y$ and the corresponding CDFs. I am looking forward to hints!

Let me show you what I have tried and where I want to get at.
Similar to $f_X$ above, I used convolution to obtain
$$f_Y (y|x) = begincases
1+y-x &mbox if y in [x-1,x] \
1-y+x &mbox if y in (x,x+1].
endcases$$
The joint density then should just be $f_X,Y(x,y) = f_X(x) f_Y(y|x)$.

My book suggests that the joint density looks like this and also that
$fracf_Y(xx)= frac2x$. Both does not coincide with what I have done. Can anyone help?

probability probability-theory probability-distributions uniform-distribution density-function

share|cite|improve this question

edited Mar 27 at 14:28

Max

asked Mar 25 at 13:10

MaxMax

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since I am very certain about my marginal density and I suppose the book's joint density is correct, I assume my conditional density appears to be incorrect. I obtained it from the marginal density of $(A-C), f_A-C$ and I plugged in $f_A-C(y-x)$. If you don't like my concrete example, I would be very happy for a general approach to determine the joint density of random variables $X,Y$ with $X= g(A,B)$ and $Y=g(C,B)$ with independent draws $A,B,C$ and some function $g$.
  $endgroup$
  – Max
  Mar 27 at 12:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "... and also that $fracx)F(x= $" What on earth does that $f(x|x)$ mean? TYpo?
  $endgroup$
  – leonbloy
  Mar 27 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, a typo. I apoligize. I edited it. It is supposed to be the conditional density over the conditional CDF evaluated at $x$.
  $endgroup$
  – Max
  Mar 27 at 14:30
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose we have three iid draws from a uniform distribution on $[0,1]$. Call these random variables $A, B$ and $C$.
Let $X=A+B$ and $Y=B+C$.
I have figured out that the density of $X$ (or $Y$) is
$$f_X(x) = begincases
x &mbox if x in [0,1] \
2-x &mbox if x in (1,2].
endcases$$
I can also see that $X=A+Y-C$.
However, I am still struggeling with the joint density of $X,Y$ and the conditional density of $X|Y$ and the corresponding CDFs. I am looking forward to hints!

Let me show you what I have tried and where I want to get at.
Similar to $f_X$ above, I used convolution to obtain
$$f_Y (y|x) = begincases
1+y-x &mbox if y in [x-1,x] \
1-y+x &mbox if y in (x,x+1].
endcases$$
The joint density then should just be $f_X,Y(x,y) = f_X(x) f_Y(y|x)$.

My book suggests that the joint density looks like this and also that
$fracf_Y(xx)= frac2x$. Both does not coincide with what I have done. Can anyone help?

probability probability-theory probability-distributions uniform-distribution density-function

share|cite|improve this question

edited Mar 27 at 14:28

Max

asked Mar 25 at 13:10

MaxMax

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since I am very certain about my marginal density and I suppose the book's joint density is correct, I assume my conditional density appears to be incorrect. I obtained it from the marginal density of $(A-C), f_A-C$ and I plugged in $f_A-C(y-x)$. If you don't like my concrete example, I would be very happy for a general approach to determine the joint density of random variables $X,Y$ with $X= g(A,B)$ and $Y=g(C,B)$ with independent draws $A,B,C$ and some function $g$.
  $endgroup$
  – Max
  Mar 27 at 12:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "... and also that $fracx)F(x= $" What on earth does that $f(x|x)$ mean? TYpo?
  $endgroup$
  – leonbloy
  Mar 27 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, a typo. I apoligize. I edited it. It is supposed to be the conditional density over the conditional CDF evaluated at $x$.
  $endgroup$
  – Max
  Mar 27 at 14:30
  

  
  
  
  

add a comment | 

$begingroup$

Suppose we have three iid draws from a uniform distribution on $[0,1]$. Call these random variables $A, B$ and $C$.
Let $X=A+B$ and $Y=B+C$.
I have figured out that the density of $X$ (or $Y$) is
$$f_X(x) = begincases
x &mbox if x in [0,1] \
2-x &mbox if x in (1,2].
endcases$$
I can also see that $X=A+Y-C$.
However, I am still struggeling with the joint density of $X,Y$ and the conditional density of $X|Y$ and the corresponding CDFs. I am looking forward to hints!

Let me show you what I have tried and where I want to get at.
Similar to $f_X$ above, I used convolution to obtain
$$f_Y (y|x) = begincases
1+y-x &mbox if y in [x-1,x] \
1-y+x &mbox if y in (x,x+1].
endcases$$
The joint density then should just be $f_X,Y(x,y) = f_X(x) f_Y(y|x)$.

My book suggests that the joint density looks like this and also that
$fracf_Y(xx)= frac2x$. Both does not coincide with what I have done. Can anyone help?

probability probability-theory probability-distributions uniform-distribution density-function

share|cite|improve this question

edited Mar 27 at 14:28

Max

asked Mar 25 at 13:10

MaxMax

62

$endgroup$

share|cite|improve this question

edited Mar 27 at 14:28

Max

asked Mar 25 at 13:10

MaxMax

62

share|cite|improve this question

edited Mar 27 at 14:28

Max

asked Mar 25 at 13:10

MaxMax

62

asked Mar 25 at 13:10

MaxMax

62

asked Mar 25 at 13:10

MaxMax

62