Convolution of weighted uniformly distributed random variables
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I have problems with deriving the CDF of a weighted difference of iid uniformly distributed random variables.
Assume that $X_1$ and $X_2$ $stackreliidsim U[0,1]$. Define Z = $acdot X_1 - X_2$, where $a > 0$. How can I derive $P(Z<z)$? I was struggling with applying the general definition of a convolution on my particular case...
Thanks in advance!
random-variables convolution uniform-distribution
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I have problems with deriving the CDF of a weighted difference of iid uniformly distributed random variables.
Assume that $X_1$ and $X_2$ $stackreliidsim U[0,1]$. Define Z = $acdot X_1 - X_2$, where $a > 0$. How can I derive $P(Z<z)$? I was struggling with applying the general definition of a convolution on my particular case...
Thanks in advance!
random-variables convolution uniform-distribution
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have problems with deriving the CDF of a weighted difference of iid uniformly distributed random variables.
Assume that $X_1$ and $X_2$ $stackreliidsim U[0,1]$. Define Z = $acdot X_1 - X_2$, where $a > 0$. How can I derive $P(Z<z)$? I was struggling with applying the general definition of a convolution on my particular case...
Thanks in advance!
random-variables convolution uniform-distribution
I have problems with deriving the CDF of a weighted difference of iid uniformly distributed random variables.
Assume that $X_1$ and $X_2$ $stackreliidsim U[0,1]$. Define Z = $acdot X_1 - X_2$, where $a > 0$. How can I derive $P(Z<z)$? I was struggling with applying the general definition of a convolution on my particular case...
Thanks in advance!
random-variables convolution uniform-distribution
asked Aug 20 at 12:46
P3rs3rk3r
385
385
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2 Answers
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On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$
Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get
$$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$
Solving the integral will give you the wanted result.
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Hint:
For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$
Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get
$$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$
Solving the integral will give you the wanted result.
add a comment |Â
up vote
0
down vote
On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$
Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get
$$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$
Solving the integral will give you the wanted result.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$
Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get
$$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$
Solving the integral will give you the wanted result.
On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = int g(x_1,x_2) f_(X_1,X_2)(x_1,x_2) dz$$ for each measurable function $g$ where $f_(X_1,X_2)$ denotes the joint density of $X_1$ and $X_2$
Now take $$A = Z < z $$ and consider that for independent random variables $X_1,X_2$ it holds $$f_(X_1,X_2)(x_1,x_2) = f_X_1(x_1)f_X_2(x_2)$$ to get
$$P(Z < z) = int_Bbb R^2 1_aX_1 - X_2 < z, dx_1dx_2$$
Solving the integral will give you the wanted result.
answered Aug 20 at 13:06
Gono
3,404316
3,404316
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up vote
0
down vote
Hint:
For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.
add a comment |Â
up vote
0
down vote
Hint:
For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.
Hint:
For fixed $z$:$$mathsf P(Z<z)=int_0^1int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $mathbb R^2tomathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.
answered Aug 20 at 13:36
drhab
87.8k541119
87.8k541119
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