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Solving Double Integral in calculating the expected value of the absolute value of the difference between two...
$begingroup$
The answer to the question is shown here, unfortunately I am having trouble evaluating the double integral shown below which is used to get the answer.
Can someone please explain how to solve the integral to get the answer shown?
$$
operatorname E[|X_1-X_2|]=int_0^2int_0^2 frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 =frac{2}{3}.
$$
Expected value of the absolute value of the difference between two independent uniform random variables?
Is the expression below correct?
$
operatorname E[|X_1-X_2|]=int_0^2int_0^{x_1} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 + int_0^2int_{x_1}^{2} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 =frac{2}{3}.
$
Can you explain how you came by the limits you suggested?
I know that for the case $E(X+Y)$ you use convolution which would give you the limits that you used. However, I would have expected that the limits would change for $|E(X-Y)|$ as the convolution is running in the opposite direction but apparently the limits stayed the same?
Tackling the integrals in order from left to right:
$
int_0^2int_0^{x_1} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1
$
$
int_0^2 frac{|x_1x_2-x_2^2/2|}{4} |_0^{x1} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2-x_1^2/2|}{4} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2|}{8} , mathrm{d}x_1
$
$
frac{|x_1^3|}{24} |_0^2
$
$
frac{|8|}{24}
$
Obviously $frac{8}{24}$ is $frac{1}{3}$. Given that the numerator is always positive the abs sign is redundant so I can just take the solution to the first part to be $frac{1}{3}$ is the reasoning correct?
$
int_0^2int_{x_1}^2 frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1
$
$
int_0^2 frac{|x_1x_2-x_2^2/2|}{4} |_{x1}^2 , mathrm{d}x_1
$
$
int_0^2 frac{|2x_1-2| - |x_1^2 - x_1^2/2|}{4} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2/2 + 2x_1 - 2|}{8} , mathrm{d}x_1
$
$
frac{|x_1^3/3 + x_1^2 - 2 x_1|}{8} |_0^2
$
$
frac{|8|}{24}
$
Again assuming I can take $frac{|8|}{24}$ to be $frac{1}{3}$ which give $frac{2}{3}$ over all.
probability probability-theory
edited Jan 28 at 15:09
Ankit Kumar
1,542221
asked Jan 28 at 13:47
BazmanBazman
406413
$endgroup$
add a comment |
$begingroup$
The answer to the question is shown here, unfortunately I am having trouble evaluating the double integral shown below which is used to get the answer.
Can someone please explain how to solve the integral to get the answer shown?
$$
operatorname E[|X_1-X_2|]=int_0^2int_0^2 frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 =frac{2}{3}.
$$
Expected value of the absolute value of the difference between two independent uniform random variables?
Is the expression below correct?
$
operatorname E[|X_1-X_2|]=int_0^2int_0^{x_1} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 + int_0^2int_{x_1}^{2} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 =frac{2}{3}.
$
Can you explain how you came by the limits you suggested?
I know that for the case $E(X+Y)$ you use convolution which would give you the limits that you used. However, I would have expected that the limits would change for $|E(X-Y)|$ as the convolution is running in the opposite direction but apparently the limits stayed the same?
Tackling the integrals in order from left to right:
$
int_0^2int_0^{x_1} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1
$
$
int_0^2 frac{|x_1x_2-x_2^2/2|}{4} |_0^{x1} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2-x_1^2/2|}{4} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2|}{8} , mathrm{d}x_1
$
$
frac{|x_1^3|}{24} |_0^2
$
$
frac{|8|}{24}
$
Obviously $frac{8}{24}$ is $frac{1}{3}$. Given that the numerator is always positive the abs sign is redundant so I can just take the solution to the first part to be $frac{1}{3}$ is the reasoning correct?
$
int_0^2int_{x_1}^2 frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1
$
$
int_0^2 frac{|x_1x_2-x_2^2/2|}{4} |_{x1}^2 , mathrm{d}x_1
$
$
int_0^2 frac{|2x_1-2| - |x_1^2 - x_1^2/2|}{4} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2/2 + 2x_1 - 2|}{8} , mathrm{d}x_1
$
$
frac{|x_1^3/3 + x_1^2 - 2 x_1|}{8} |_0^2
$
$
frac{|8|}{24}
$
Again assuming I can take $frac{|8|}{24}$ to be $frac{1}{3}$ which give $frac{2}{3}$ over all.
probability probability-theory
edited Jan 28 at 15:09
Ankit Kumar
1,542221
asked Jan 28 at 13:47
BazmanBazman
406413
$endgroup$
1
$begingroup$
Try dividing the $dx_2$ integral on two integrals {$[0,x_1], [x_1,2]$}
$endgroup$
– vermator
Jan 28 at 13:59
add a comment |
$begingroup$
The answer to the question is shown here, unfortunately I am having trouble evaluating the double integral shown below which is used to get the answer.
Can someone please explain how to solve the integral to get the answer shown?
$$
operatorname E[|X_1-X_2|]=int_0^2int_0^2 frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 =frac{2}{3}.
$$
Expected value of the absolute value of the difference between two independent uniform random variables?
Is the expression below correct?
$
operatorname E[|X_1-X_2|]=int_0^2int_0^{x_1} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 + int_0^2int_{x_1}^{2} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 =frac{2}{3}.
$
Can you explain how you came by the limits you suggested?
I know that for the case $E(X+Y)$ you use convolution which would give you the limits that you used. However, I would have expected that the limits would change for $|E(X-Y)|$ as the convolution is running in the opposite direction but apparently the limits stayed the same?
Tackling the integrals in order from left to right:
$
int_0^2int_0^{x_1} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1
$
$
int_0^2 frac{|x_1x_2-x_2^2/2|}{4} |_0^{x1} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2-x_1^2/2|}{4} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2|}{8} , mathrm{d}x_1
$
$
frac{|x_1^3|}{24} |_0^2
$
$
frac{|8|}{24}
$
Obviously $frac{8}{24}$ is $frac{1}{3}$. Given that the numerator is always positive the abs sign is redundant so I can just take the solution to the first part to be $frac{1}{3}$ is the reasoning correct?
$
int_0^2int_{x_1}^2 frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1
$
$
int_0^2 frac{|x_1x_2-x_2^2/2|}{4} |_{x1}^2 , mathrm{d}x_1
$
$
int_0^2 frac{|2x_1-2| - |x_1^2 - x_1^2/2|}{4} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2/2 + 2x_1 - 2|}{8} , mathrm{d}x_1
$
$
frac{|x_1^3/3 + x_1^2 - 2 x_1|}{8} |_0^2
$
$
frac{|8|}{24}
$
Again assuming I can take $frac{|8|}{24}$ to be $frac{1}{3}$ which give $frac{2}{3}$ over all.
probability probability-theory
edited Jan 28 at 15:09
Ankit Kumar
1,542221
asked Jan 28 at 13:47
BazmanBazman
406413
$endgroup$
The answer to the question is shown here, unfortunately I am having trouble evaluating the double integral shown below which is used to get the answer.
Can someone please explain how to solve the integral to get the answer shown?
$$
operatorname E[|X_1-X_2|]=int_0^2int_0^2 frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 =frac{2}{3}.
$$
Expected value of the absolute value of the difference between two independent uniform random variables?
Is the expression below correct?
$
operatorname E[|X_1-X_2|]=int_0^2int_0^{x_1} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 + int_0^2int_{x_1}^{2} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1 =frac{2}{3}.
$
Can you explain how you came by the limits you suggested?
I know that for the case $E(X+Y)$ you use convolution which would give you the limits that you used. However, I would have expected that the limits would change for $|E(X-Y)|$ as the convolution is running in the opposite direction but apparently the limits stayed the same?
Tackling the integrals in order from left to right:
$
int_0^2int_0^{x_1} frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1
$
$
int_0^2 frac{|x_1x_2-x_2^2/2|}{4} |_0^{x1} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2-x_1^2/2|}{4} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2|}{8} , mathrm{d}x_1
$
$
frac{|x_1^3|}{24} |_0^2
$
$
frac{|8|}{24}
$
Obviously $frac{8}{24}$ is $frac{1}{3}$. Given that the numerator is always positive the abs sign is redundant so I can just take the solution to the first part to be $frac{1}{3}$ is the reasoning correct?
$
int_0^2int_{x_1}^2 frac{|x_1-x_2|}{4} , mathrm{d}x_2 , mathrm{d}x_1
$
$
int_0^2 frac{|x_1x_2-x_2^2/2|}{4} |_{x1}^2 , mathrm{d}x_1
$
$
int_0^2 frac{|2x_1-2| - |x_1^2 - x_1^2/2|}{4} , mathrm{d}x_1
$
$
int_0^2 frac{|x_1^2/2 + 2x_1 - 2|}{8} , mathrm{d}x_1
$
$
frac{|x_1^3/3 + x_1^2 - 2 x_1|}{8} |_0^2
$
$
frac{|8|}{24}
$
Again assuming I can take $frac{|8|}{24}$ to be $frac{1}{3}$ which give $frac{2}{3}$ over all.
probability probability-theory
probability probability-theory
edited Jan 28 at 15:09
Ankit Kumar
1,542221
asked Jan 28 at 13:47
BazmanBazman
406413
edited Jan 28 at 15:09
Ankit Kumar
1,542221
asked Jan 28 at 13:47
BazmanBazman
406413
edited Jan 28 at 15:09
Ankit Kumar
1,542221
edited Jan 28 at 15:09
Ankit Kumar
1,542221
edited Jan 28 at 15:09
Ankit Kumar
1,542221
1,542221
asked Jan 28 at 13:47
BazmanBazman
406413
asked Jan 28 at 13:47
BazmanBazman
406413
asked Jan 28 at 13:47
BazmanBazman
406413
406413
1
$begingroup$
Try dividing the $dx_2$ integral on two integrals {$[0,x_1], [x_1,2]$}
$endgroup$
– vermator
Jan 28 at 13:59
add a comment |
1
$begingroup$
Try dividing the $dx_2$ integral on two integrals {$[0,x_1], [x_1,2]$}
$endgroup$
– vermator
Jan 28 at 13:59
1
1
$begingroup$
Try dividing the $dx_2$ integral on two integrals {$[0,x_1], [x_1,2]$}
$endgroup$
– vermator
Jan 28 at 13:59
$begingroup$
Try dividing the $dx_2$ integral on two integrals {$[0,x_1], [x_1,2]$}
$endgroup$
– vermator
Jan 28 at 13:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $X_1$ and $X_2$ are uniformily distributed from U(0,2),
then $Z = |X_1-X_2|$ has a pdf of
$P(Z=z) = frac{1}{2}(2-z)$, $0le z le 2$
Thus $E(Z) = int_{0}^{2} frac{1}{2}z(2-z)dz$
$=frac{2}{3}$
Easy Understanding of Convolution The best way to understand convolution is given in the article in the link,using that
I am going to solve the above problem and hence you could follow the same for any similar problem such as this with not too much confusion.
$Z = X-Y$ where X and Y are U(0,2).
I am going to define a new variable W where W is distributed according to U(-2,0)
Thus $Z = X - Y = X+ W$ where X is U(0,1) and W is U(-2,0).
Now I am going define the bounds
$t_{X_0} = -2$
$t_{X_1} = 0$
$t_{W_0} = 0$
$t_{W_1} = 2$
Thus $$f_Z(z) = 0, z le t_{X_0}+t_{W_0} ,$$
$$f_Z(z) = int_{max(t_{W_0}, t-t_{X_1})}^{min(t_{W_1}, t-t_{X_0})} f_W(w)f_X(z-w)dw, text{ } t_{X_0}+t_{W_0} le z le t_{X_1}+t_{W_1},$$
$$f_Z(z) = 0, z ge t_{X_1}+t_{W_1} ,$$
These translate to the following:
$$f_Z(z) = 0, z le -2 $$
$$f_Z(z) = int_{max(0, z)}^{min(2, z+2)} f_W(w)f_X(z-w)dw, text{ } -2le z le 2,$$
$$f_Z(z) = 0, z ge 2 ,$$
$f_W(w) = frac{1}{2}$ as $W$ is $U(-2,0)$.
$f_X(x) = frac{1}{2} $ as $X$ is $U(0,2)$,
The middle one needs to be split into two intervals, and they are a) $-2le zle 0$, b) $0le zle 2$.
Thus
$f_Z(z) = int_{0}^{z+2}frac{1}{4}dw = frac{z+2}{4}$, $-2le zle 0$
$f_Z(z) = int_{z}^{2}frac{1}{4}dw = frac{2-z}{4}$, $0le zle 2$
Sanity check is to find if $int_{-2}^{2} f_Z(z) = 1$ which it is in this case You have to find the pdf of Z = |X-Y| and the only interval for this is $0le zle 2$ and the pdf is twice that of $2frac{2-z}{4} = frac{2-z}{2}$
and hence
The pdf $boxed{f_Z(z) = frac{2-z}{2}, 0le z le 2}$
Goodluck
answered Jan 28 at 16:02
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
Can you elaborate on how you calculated the pdf please.
$endgroup$
– Bazman
Jan 28 at 16:57
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
If $X_1$ and $X_2$ are uniformily distributed from U(0,2),
then $Z = |X_1-X_2|$ has a pdf of
$P(Z=z) = frac{1}{2}(2-z)$, $0le z le 2$
Thus $E(Z) = int_{0}^{2} frac{1}{2}z(2-z)dz$
$=frac{2}{3}$
Easy Understanding of Convolution The best way to understand convolution is given in the article in the link,using that
I am going to solve the above problem and hence you could follow the same for any similar problem such as this with not too much confusion.
$Z = X-Y$ where X and Y are U(0,2).
I am going to define a new variable W where W is distributed according to U(-2,0)
Thus $Z = X - Y = X+ W$ where X is U(0,1) and W is U(-2,0).
Now I am going define the bounds
$t_{X_0} = -2$
$t_{X_1} = 0$
$t_{W_0} = 0$
$t_{W_1} = 2$
Thus $$f_Z(z) = 0, z le t_{X_0}+t_{W_0} ,$$
$$f_Z(z) = int_{max(t_{W_0}, t-t_{X_1})}^{min(t_{W_1}, t-t_{X_0})} f_W(w)f_X(z-w)dw, text{ } t_{X_0}+t_{W_0} le z le t_{X_1}+t_{W_1},$$
$$f_Z(z) = 0, z ge t_{X_1}+t_{W_1} ,$$
These translate to the following:
$$f_Z(z) = 0, z le -2 $$
$$f_Z(z) = int_{max(0, z)}^{min(2, z+2)} f_W(w)f_X(z-w)dw, text{ } -2le z le 2,$$
$$f_Z(z) = 0, z ge 2 ,$$
$f_W(w) = frac{1}{2}$ as $W$ is $U(-2,0)$.
$f_X(x) = frac{1}{2} $ as $X$ is $U(0,2)$,
The middle one needs to be split into two intervals, and they are a) $-2le zle 0$, b) $0le zle 2$.
Thus
$f_Z(z) = int_{0}^{z+2}frac{1}{4}dw = frac{z+2}{4}$, $-2le zle 0$
$f_Z(z) = int_{z}^{2}frac{1}{4}dw = frac{2-z}{4}$, $0le zle 2$
Sanity check is to find if $int_{-2}^{2} f_Z(z) = 1$ which it is in this case You have to find the pdf of Z = |X-Y| and the only interval for this is $0le zle 2$ and the pdf is twice that of $2frac{2-z}{4} = frac{2-z}{2}$
and hence
The pdf $boxed{f_Z(z) = frac{2-z}{2}, 0le z le 2}$
Goodluck
answered Jan 28 at 16:02
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
Can you elaborate on how you calculated the pdf please.
$endgroup$
– Bazman
Jan 28 at 16:57
add a comment |
$begingroup$
If $X_1$ and $X_2$ are uniformily distributed from U(0,2),
then $Z = |X_1-X_2|$ has a pdf of
$P(Z=z) = frac{1}{2}(2-z)$, $0le z le 2$
Thus $E(Z) = int_{0}^{2} frac{1}{2}z(2-z)dz$
$=frac{2}{3}$
Easy Understanding of Convolution The best way to understand convolution is given in the article in the link,using that
I am going to solve the above problem and hence you could follow the same for any similar problem such as this with not too much confusion.
$Z = X-Y$ where X and Y are U(0,2).
I am going to define a new variable W where W is distributed according to U(-2,0)
Thus $Z = X - Y = X+ W$ where X is U(0,1) and W is U(-2,0).
Now I am going define the bounds
$t_{X_0} = -2$
$t_{X_1} = 0$
$t_{W_0} = 0$
$t_{W_1} = 2$
Thus $$f_Z(z) = 0, z le t_{X_0}+t_{W_0} ,$$
$$f_Z(z) = int_{max(t_{W_0}, t-t_{X_1})}^{min(t_{W_1}, t-t_{X_0})} f_W(w)f_X(z-w)dw, text{ } t_{X_0}+t_{W_0} le z le t_{X_1}+t_{W_1},$$
$$f_Z(z) = 0, z ge t_{X_1}+t_{W_1} ,$$
These translate to the following:
$$f_Z(z) = 0, z le -2 $$
$$f_Z(z) = int_{max(0, z)}^{min(2, z+2)} f_W(w)f_X(z-w)dw, text{ } -2le z le 2,$$
$$f_Z(z) = 0, z ge 2 ,$$
$f_W(w) = frac{1}{2}$ as $W$ is $U(-2,0)$.
$f_X(x) = frac{1}{2} $ as $X$ is $U(0,2)$,
The middle one needs to be split into two intervals, and they are a) $-2le zle 0$, b) $0le zle 2$.
Thus
$f_Z(z) = int_{0}^{z+2}frac{1}{4}dw = frac{z+2}{4}$, $-2le zle 0$
$f_Z(z) = int_{z}^{2}frac{1}{4}dw = frac{2-z}{4}$, $0le zle 2$
Sanity check is to find if $int_{-2}^{2} f_Z(z) = 1$ which it is in this case You have to find the pdf of Z = |X-Y| and the only interval for this is $0le zle 2$ and the pdf is twice that of $2frac{2-z}{4} = frac{2-z}{2}$
and hence
The pdf $boxed{f_Z(z) = frac{2-z}{2}, 0le z le 2}$
Goodluck
answered Jan 28 at 16:02
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
Can you elaborate on how you calculated the pdf please.
$endgroup$
– Bazman
Jan 28 at 16:57
add a comment |
$begingroup$
If $X_1$ and $X_2$ are uniformily distributed from U(0,2),
then $Z = |X_1-X_2|$ has a pdf of
$P(Z=z) = frac{1}{2}(2-z)$, $0le z le 2$
Thus $E(Z) = int_{0}^{2} frac{1}{2}z(2-z)dz$
$=frac{2}{3}$
Easy Understanding of Convolution The best way to understand convolution is given in the article in the link,using that
I am going to solve the above problem and hence you could follow the same for any similar problem such as this with not too much confusion.
$Z = X-Y$ where X and Y are U(0,2).
I am going to define a new variable W where W is distributed according to U(-2,0)
Thus $Z = X - Y = X+ W$ where X is U(0,1) and W is U(-2,0).
Now I am going define the bounds
$t_{X_0} = -2$
$t_{X_1} = 0$
$t_{W_0} = 0$
$t_{W_1} = 2$
Thus $$f_Z(z) = 0, z le t_{X_0}+t_{W_0} ,$$
$$f_Z(z) = int_{max(t_{W_0}, t-t_{X_1})}^{min(t_{W_1}, t-t_{X_0})} f_W(w)f_X(z-w)dw, text{ } t_{X_0}+t_{W_0} le z le t_{X_1}+t_{W_1},$$
$$f_Z(z) = 0, z ge t_{X_1}+t_{W_1} ,$$
These translate to the following:
$$f_Z(z) = 0, z le -2 $$
$$f_Z(z) = int_{max(0, z)}^{min(2, z+2)} f_W(w)f_X(z-w)dw, text{ } -2le z le 2,$$
$$f_Z(z) = 0, z ge 2 ,$$
$f_W(w) = frac{1}{2}$ as $W$ is $U(-2,0)$.
$f_X(x) = frac{1}{2} $ as $X$ is $U(0,2)$,
The middle one needs to be split into two intervals, and they are a) $-2le zle 0$, b) $0le zle 2$.
Thus
$f_Z(z) = int_{0}^{z+2}frac{1}{4}dw = frac{z+2}{4}$, $-2le zle 0$
$f_Z(z) = int_{z}^{2}frac{1}{4}dw = frac{2-z}{4}$, $0le zle 2$
Sanity check is to find if $int_{-2}^{2} f_Z(z) = 1$ which it is in this case You have to find the pdf of Z = |X-Y| and the only interval for this is $0le zle 2$ and the pdf is twice that of $2frac{2-z}{4} = frac{2-z}{2}$
and hence
The pdf $boxed{f_Z(z) = frac{2-z}{2}, 0le z le 2}$
Goodluck
answered Jan 28 at 16:02
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
If $X_1$ and $X_2$ are uniformily distributed from U(0,2),
then $Z = |X_1-X_2|$ has a pdf of
$P(Z=z) = frac{1}{2}(2-z)$, $0le z le 2$
Thus $E(Z) = int_{0}^{2} frac{1}{2}z(2-z)dz$
$=frac{2}{3}$
Easy Understanding of Convolution The best way to understand convolution is given in the article in the link,using that
I am going to solve the above problem and hence you could follow the same for any similar problem such as this with not too much confusion.
$Z = X-Y$ where X and Y are U(0,2).
I am going to define a new variable W where W is distributed according to U(-2,0)
Thus $Z = X - Y = X+ W$ where X is U(0,1) and W is U(-2,0).
Now I am going define the bounds
$t_{X_0} = -2$
$t_{X_1} = 0$
$t_{W_0} = 0$
$t_{W_1} = 2$
Thus $$f_Z(z) = 0, z le t_{X_0}+t_{W_0} ,$$
$$f_Z(z) = int_{max(t_{W_0}, t-t_{X_1})}^{min(t_{W_1}, t-t_{X_0})} f_W(w)f_X(z-w)dw, text{ } t_{X_0}+t_{W_0} le z le t_{X_1}+t_{W_1},$$
$$f_Z(z) = 0, z ge t_{X_1}+t_{W_1} ,$$
These translate to the following:
$$f_Z(z) = 0, z le -2 $$
$$f_Z(z) = int_{max(0, z)}^{min(2, z+2)} f_W(w)f_X(z-w)dw, text{ } -2le z le 2,$$
$$f_Z(z) = 0, z ge 2 ,$$
$f_W(w) = frac{1}{2}$ as $W$ is $U(-2,0)$.
$f_X(x) = frac{1}{2} $ as $X$ is $U(0,2)$,
The middle one needs to be split into two intervals, and they are a) $-2le zle 0$, b) $0le zle 2$.
Thus
$f_Z(z) = int_{0}^{z+2}frac{1}{4}dw = frac{z+2}{4}$, $-2le zle 0$
$f_Z(z) = int_{z}^{2}frac{1}{4}dw = frac{2-z}{4}$, $0le zle 2$
Sanity check is to find if $int_{-2}^{2} f_Z(z) = 1$ which it is in this case You have to find the pdf of Z = |X-Y| and the only interval for this is $0le zle 2$ and the pdf is twice that of $2frac{2-z}{4} = frac{2-z}{2}$
and hence
The pdf $boxed{f_Z(z) = frac{2-z}{2}, 0le z le 2}$
Goodluck
answered Jan 28 at 16:02
Satish RamanathanSatish Ramanathan
10k31323
edited Jan 29 at 15:02
answered Jan 28 at 16:02
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 28 at 16:02
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 28 at 16:02
Satish RamanathanSatish Ramanathan
10k31323
10k31323
$begingroup$
Can you elaborate on how you calculated the pdf please.
$endgroup$
– Bazman
Jan 28 at 16:57
add a comment |
$begingroup$
Can you elaborate on how you calculated the pdf please.
$endgroup$
– Bazman
Jan 28 at 16:57
$begingroup$
Can you elaborate on how you calculated the pdf please.
$endgroup$
– Bazman
Jan 28 at 16:57
$begingroup$
Can you elaborate on how you calculated the pdf please.
$endgroup$
– Bazman
Jan 28 at 16:57
add a comment |
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$begingroup$
Try dividing the $dx_2$ integral on two integrals {$[0,x_1], [x_1,2]$}
$endgroup$
– vermator
Jan 28 at 13:59