Z=X+Y pdf, there is a part I can't understand.
Two independent random variable X,Y which is U(0,1)
what is the pdf of Z=X+Y?
$F_Z(z)$ =
begin{cases}
0, & zle0 \
int_0^z int_0^{z-y} 1 ,dxdy=z^2/2, & 0lt zle1 \
1-int_bbox[yellow]{z-1}^1 int_{z-y}^1 1 ,dxdy=1-(2-z)^2/2,& 1lt zle2 \
1 & zge2
end{cases}
I understand I have to differential $F_z(z)$
The thing I can't understand is the highlight part.
Why does it starts with z-1?
statistics random-variables
asked 2 days ago
yonghankwon0yonghankwon0
134
add a comment |
Two independent random variable X,Y which is U(0,1)
what is the pdf of Z=X+Y?
$F_Z(z)$ =
begin{cases}
0, & zle0 \
int_0^z int_0^{z-y} 1 ,dxdy=z^2/2, & 0lt zle1 \
1-int_bbox[yellow]{z-1}^1 int_{z-y}^1 1 ,dxdy=1-(2-z)^2/2,& 1lt zle2 \
1 & zge2
end{cases}
I understand I have to differential $F_z(z)$
The thing I can't understand is the highlight part.
Why does it starts with z-1?
statistics random-variables
asked 2 days ago
yonghankwon0yonghankwon0
134
1
By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
– J.G.
2 days ago
@J.G. Thank you I edited it.
– yonghankwon0
yesterday
add a comment |
Two independent random variable X,Y which is U(0,1)
what is the pdf of Z=X+Y?
$F_Z(z)$ =
begin{cases}
0, & zle0 \
int_0^z int_0^{z-y} 1 ,dxdy=z^2/2, & 0lt zle1 \
1-int_bbox[yellow]{z-1}^1 int_{z-y}^1 1 ,dxdy=1-(2-z)^2/2,& 1lt zle2 \
1 & zge2
end{cases}
I understand I have to differential $F_z(z)$
The thing I can't understand is the highlight part.
Why does it starts with z-1?
statistics random-variables
asked 2 days ago
yonghankwon0yonghankwon0
134
Two independent random variable X,Y which is U(0,1)
what is the pdf of Z=X+Y?
$F_Z(z)$ =
begin{cases}
0, & zle0 \
int_0^z int_0^{z-y} 1 ,dxdy=z^2/2, & 0lt zle1 \
1-int_bbox[yellow]{z-1}^1 int_{z-y}^1 1 ,dxdy=1-(2-z)^2/2,& 1lt zle2 \
1 & zge2
end{cases}
I understand I have to differential $F_z(z)$
The thing I can't understand is the highlight part.
Why does it starts with z-1?
statistics random-variables
statistics random-variables
asked 2 days ago
yonghankwon0yonghankwon0
134
asked 2 days ago
yonghankwon0yonghankwon0
134
edited yesterday
yonghankwon0
asked 2 days ago
yonghankwon0yonghankwon0
134
asked 2 days ago
yonghankwon0yonghankwon0
134
asked 2 days ago
yonghankwon0yonghankwon0
134
134
1
By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
– J.G.
2 days ago
@J.G. Thank you I edited it.
– yonghankwon0
yesterday
add a comment |
1
By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
– J.G.
2 days ago
@J.G. Thank you I edited it.
– yonghankwon0
yesterday
1
1
By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
– J.G.
2 days ago
By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
– J.G.
2 days ago
@J.G. Thank you I edited it.
– yonghankwon0
yesterday
@J.G. Thank you I edited it.
– yonghankwon0
yesterday
add a comment |
1 Answer
1
active
oldest
votes
Convolution of probability densities
Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
Define $Z equiv X + Y$.
Note that
$$
F_Z(z)
= mathbb{P}(Z leq z)
= mathbb{P}(X + Y leq z)
= int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
= int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
$$
Differentiating,
$$
f_Z(z)
= int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
$$
Convolution of i.i.d. uniform distributions
If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
Plugging this into the integral of the previous section,
$$
f_Z(z)
= int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
$$
Proceeding by cases,
$$
f_Z(z)
= begin{cases}
int_0^z dx = z & text{if } 0 leq z leq 1 \
int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
end{cases}
$$
answered yesterday
parsiadparsiad
16.9k32353
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Convolution of probability densities
Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
Define $Z equiv X + Y$.
Note that
$$
F_Z(z)
= mathbb{P}(Z leq z)
= mathbb{P}(X + Y leq z)
= int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
= int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
$$
Differentiating,
$$
f_Z(z)
= int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
$$
Convolution of i.i.d. uniform distributions
If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
Plugging this into the integral of the previous section,
$$
f_Z(z)
= int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
$$
Proceeding by cases,
$$
f_Z(z)
= begin{cases}
int_0^z dx = z & text{if } 0 leq z leq 1 \
int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
end{cases}
$$
answered yesterday
parsiadparsiad
16.9k32353
add a comment |
Convolution of probability densities
Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
Define $Z equiv X + Y$.
Note that
$$
F_Z(z)
= mathbb{P}(Z leq z)
= mathbb{P}(X + Y leq z)
= int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
= int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
$$
Differentiating,
$$
f_Z(z)
= int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
$$
Convolution of i.i.d. uniform distributions
If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
Plugging this into the integral of the previous section,
$$
f_Z(z)
= int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
$$
Proceeding by cases,
$$
f_Z(z)
= begin{cases}
int_0^z dx = z & text{if } 0 leq z leq 1 \
int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
end{cases}
$$
answered yesterday
parsiadparsiad
16.9k32353
add a comment |
Convolution of probability densities
Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
Define $Z equiv X + Y$.
Note that
$$
F_Z(z)
= mathbb{P}(Z leq z)
= mathbb{P}(X + Y leq z)
= int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
= int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
$$
Differentiating,
$$
f_Z(z)
= int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
$$
Convolution of i.i.d. uniform distributions
If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
Plugging this into the integral of the previous section,
$$
f_Z(z)
= int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
$$
Proceeding by cases,
$$
f_Z(z)
= begin{cases}
int_0^z dx = z & text{if } 0 leq z leq 1 \
int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
end{cases}
$$
answered yesterday
parsiadparsiad
16.9k32353
Convolution of probability densities
Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$.
Define $Z equiv X + Y$.
Note that
$$
F_Z(z)
= mathbb{P}(Z leq z)
= mathbb{P}(X + Y leq z)
= int_{-infty}^{infty} f_X(x) mathbb{P}(x + Y leq z) dx
= int_{-infty}^{infty} f_X(x) F_Y(z - x) dx.
$$
Differentiating,
$$
f_Z(z)
= int_{-infty}^{infty} f_X(x) f_Y(z - x) dx.
$$
Convolution of i.i.d. uniform distributions
If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = boldsymbol{1}_{(0,1)}$ where $boldsymbol{1}$ is the indicator function.
Plugging this into the integral of the previous section,
$$
f_Z(z)
= int_0^1 boldsymbol{1}_{(0,1)}(z - x) dx.
$$
Proceeding by cases,
$$
f_Z(z)
= begin{cases}
int_0^z dx = z & text{if } 0 leq z leq 1 \
int_bbox[yellow]{z-1}^1 dx = 2 - z & text{if } 1 leq z leq 2. \
end{cases}
$$
answered yesterday
parsiadparsiad
16.9k32353
answered yesterday
parsiadparsiad
16.9k32353
answered yesterday
parsiadparsiad
16.9k32353
answered yesterday
parsiadparsiad
16.9k32353
16.9k32353
add a comment |
add a comment |
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1
By union, did you mean uniform? The usual notation for this distribution is $U(0,,1)$.
– J.G.
2 days ago
@J.G. Thank you I edited it.
– yonghankwon0
yesterday