Computing the PDF of $X + Y$ if $X, Y sim text{U}(0, 1)$ using a convolution
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Theorem: If $X$ and $Y$ are independent continuous random variables with density functions $f_{X}$ and $f_{Y}$, the probability
distribution $f_{X + Y}$ of $X + Y$ is given by
$$f_{X + Y}(a) = int_{-infty}^{infty} f_{X}(a - y) f_{Y}(y)
mathop{dy} $$
If $X$ and $Y$ are independent random variables, both uniformly distributed on $(0, 1)$, calculate the probability density of $X + Y$.
We have
$$f_{X}(a) = f_{Y}(a) = begin{cases}
1 & text{ if } 0 < a < 1 \
0 & text{ otherwise.}
end{cases} $$
So, we obtain
$$f_{X + Y}(a) = int_{-infty}^{infty} f_{X}(a - y)f_{Y}(y) mathop{dy}. $$
$$= int_{0}^{1} f_{X}(a - y) mathop{dy}. $$
(My first question here is what happened to the $f_{Y}(y)$ term? Did they just equate it to $1$ because a probability distribution has total area $1$ under the curve? Wouldn't this be assuming that the integral of the product of two functions is equal to the product of their integrals?).
Okay, from here, the book writes for $0 leq a leq 1$, this yields
$$f_{X + Y}(a) = int_{0}^{a} mathop{dy} = a $$
and for $1 < a < 2$, we get
$$f_{X + Y}(a) = int_{a - 1}^{1} mathop{dy} = 2 - a. $$
I am also lost here. How did they figure out to split it into two cases, and how did they know it is on the interval $0$ to $2$?
probability probability-theory probability-distributions convolution uniform-distribution
edited Nov 26 at 15:21
Davide Giraudo
124k16150259
asked Nov 26 at 6:04
stackofhay42
1424
add a comment |
up vote
1
down vote
favorite
Theorem: If $X$ and $Y$ are independent continuous random variables with density functions $f_{X}$ and $f_{Y}$, the probability
distribution $f_{X + Y}$ of $X + Y$ is given by
$$f_{X + Y}(a) = int_{-infty}^{infty} f_{X}(a - y) f_{Y}(y)
mathop{dy} $$
If $X$ and $Y$ are independent random variables, both uniformly distributed on $(0, 1)$, calculate the probability density of $X + Y$.
We have
$$f_{X}(a) = f_{Y}(a) = begin{cases}
1 & text{ if } 0 < a < 1 \
0 & text{ otherwise.}
end{cases} $$
So, we obtain
$$f_{X + Y}(a) = int_{-infty}^{infty} f_{X}(a - y)f_{Y}(y) mathop{dy}. $$
$$= int_{0}^{1} f_{X}(a - y) mathop{dy}. $$
(My first question here is what happened to the $f_{Y}(y)$ term? Did they just equate it to $1$ because a probability distribution has total area $1$ under the curve? Wouldn't this be assuming that the integral of the product of two functions is equal to the product of their integrals?).
Okay, from here, the book writes for $0 leq a leq 1$, this yields
$$f_{X + Y}(a) = int_{0}^{a} mathop{dy} = a $$
and for $1 < a < 2$, we get
$$f_{X + Y}(a) = int_{a - 1}^{1} mathop{dy} = 2 - a. $$
I am also lost here. How did they figure out to split it into two cases, and how did they know it is on the interval $0$ to $2$?
probability probability-theory probability-distributions convolution uniform-distribution
edited Nov 26 at 15:21
Davide Giraudo
124k16150259
asked Nov 26 at 6:04
stackofhay42
1424
1
The density of $f_Y$ is $1$ so there's no need to write it. It should be clear that adding two $(0, 1)$ variables means the sum can be at most $2.$ Also, uniform plus uniform is triangle.
– Sean Roberson
Nov 26 at 6:12
how'd they know to split it into the case where $0 leq a leq 1$ and $1 < a < 2$? why not like $0 leq a leq 1/2$ etc?
– stackofhay42
Nov 26 at 6:13
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Theorem: If $X$ and $Y$ are independent continuous random variables with density functions $f_{X}$ and $f_{Y}$, the probability
distribution $f_{X + Y}$ of $X + Y$ is given by
$$f_{X + Y}(a) = int_{-infty}^{infty} f_{X}(a - y) f_{Y}(y)
mathop{dy} $$
If $X$ and $Y$ are independent random variables, both uniformly distributed on $(0, 1)$, calculate the probability density of $X + Y$.
We have
$$f_{X}(a) = f_{Y}(a) = begin{cases}
1 & text{ if } 0 < a < 1 \
0 & text{ otherwise.}
end{cases} $$
So, we obtain
$$f_{X + Y}(a) = int_{-infty}^{infty} f_{X}(a - y)f_{Y}(y) mathop{dy}. $$
$$= int_{0}^{1} f_{X}(a - y) mathop{dy}. $$
(My first question here is what happened to the $f_{Y}(y)$ term? Did they just equate it to $1$ because a probability distribution has total area $1$ under the curve? Wouldn't this be assuming that the integral of the product of two functions is equal to the product of their integrals?).
Okay, from here, the book writes for $0 leq a leq 1$, this yields
$$f_{X + Y}(a) = int_{0}^{a} mathop{dy} = a $$
and for $1 < a < 2$, we get
$$f_{X + Y}(a) = int_{a - 1}^{1} mathop{dy} = 2 - a. $$
I am also lost here. How did they figure out to split it into two cases, and how did they know it is on the interval $0$ to $2$?
probability probability-theory probability-distributions convolution uniform-distribution
edited Nov 26 at 15:21
Davide Giraudo
124k16150259
asked Nov 26 at 6:04
stackofhay42
1424
Theorem: If $X$ and $Y$ are independent continuous random variables with density functions $f_{X}$ and $f_{Y}$, the probability
distribution $f_{X + Y}$ of $X + Y$ is given by
$$f_{X + Y}(a) = int_{-infty}^{infty} f_{X}(a - y) f_{Y}(y)
mathop{dy} $$
If $X$ and $Y$ are independent random variables, both uniformly distributed on $(0, 1)$, calculate the probability density of $X + Y$.
We have
$$f_{X}(a) = f_{Y}(a) = begin{cases}
1 & text{ if } 0 < a < 1 \
0 & text{ otherwise.}
end{cases} $$
So, we obtain
$$f_{X + Y}(a) = int_{-infty}^{infty} f_{X}(a - y)f_{Y}(y) mathop{dy}. $$
$$= int_{0}^{1} f_{X}(a - y) mathop{dy}. $$
(My first question here is what happened to the $f_{Y}(y)$ term? Did they just equate it to $1$ because a probability distribution has total area $1$ under the curve? Wouldn't this be assuming that the integral of the product of two functions is equal to the product of their integrals?).
Okay, from here, the book writes for $0 leq a leq 1$, this yields
$$f_{X + Y}(a) = int_{0}^{a} mathop{dy} = a $$
and for $1 < a < 2$, we get
$$f_{X + Y}(a) = int_{a - 1}^{1} mathop{dy} = 2 - a. $$
I am also lost here. How did they figure out to split it into two cases, and how did they know it is on the interval $0$ to $2$?
probability probability-theory probability-distributions convolution uniform-distribution
probability probability-theory probability-distributions convolution uniform-distribution
edited Nov 26 at 15:21
Davide Giraudo
124k16150259
asked Nov 26 at 6:04
stackofhay42
1424
edited Nov 26 at 15:21
Davide Giraudo
124k16150259
asked Nov 26 at 6:04
stackofhay42
1424
edited Nov 26 at 15:21
Davide Giraudo
124k16150259
edited Nov 26 at 15:21
Davide Giraudo
124k16150259
edited Nov 26 at 15:21
Davide Giraudo
124k16150259
124k16150259
asked Nov 26 at 6:04
stackofhay42
1424
asked Nov 26 at 6:04
stackofhay42
1424
asked Nov 26 at 6:04
stackofhay42
1424
1424
1
The density of $f_Y$ is $1$ so there's no need to write it. It should be clear that adding two $(0, 1)$ variables means the sum can be at most $2.$ Also, uniform plus uniform is triangle.
– Sean Roberson
Nov 26 at 6:12
how'd they know to split it into the case where $0 leq a leq 1$ and $1 < a < 2$? why not like $0 leq a leq 1/2$ etc?
– stackofhay42
Nov 26 at 6:13
add a comment |
1
The density of $f_Y$ is $1$ so there's no need to write it. It should be clear that adding two $(0, 1)$ variables means the sum can be at most $2.$ Also, uniform plus uniform is triangle.
– Sean Roberson
Nov 26 at 6:12
how'd they know to split it into the case where $0 leq a leq 1$ and $1 < a < 2$? why not like $0 leq a leq 1/2$ etc?
– stackofhay42
Nov 26 at 6:13
1
1
The density of $f_Y$ is $1$ so there's no need to write it. It should be clear that adding two $(0, 1)$ variables means the sum can be at most $2.$ Also, uniform plus uniform is triangle.
– Sean Roberson
Nov 26 at 6:12
The density of $f_Y$ is $1$ so there's no need to write it. It should be clear that adding two $(0, 1)$ variables means the sum can be at most $2.$ Also, uniform plus uniform is triangle.
– Sean Roberson
Nov 26 at 6:12
how'd they know to split it into the case where $0 leq a leq 1$ and $1 < a < 2$? why not like $0 leq a leq 1/2$ etc?
– stackofhay42
Nov 26 at 6:13
how'd they know to split it into the case where $0 leq a leq 1$ and $1 < a < 2$? why not like $0 leq a leq 1/2$ etc?
– stackofhay42
Nov 26 at 6:13
add a comment |
1 Answer
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Answer to your first question: As you yourself noted, $f_Y(a) = 1 ; forall a in [0,1]$, so the $f_Y(a)$ term disappears from the integration. It is not because $int_0^1 f_Y(a)da=1$, which is also true. Also, the limits of integration changed from $(-infty,infty)$ to $[0,1]$ because $f_Y(a)=0$ outside $[0,1]$.
Now you are left with the term $f_X(a-y)$, which equals 1 when $(a-y) in [0,1]$ and 0 otherwise. If $a leq 1$, this holds true when $y leq a$ and $a-y leq 1$. The latter condition can be rewritten as $y geq a-1$. Since $a leq 1$, the condition is equivalent to $y geq 0$. Hence, the limits of integration are $[0,a]$ and over this range $f_x(a-y)=1$, by definition.
By similar reasoning, if $a>1$, $f_X(a-y)=1$ only when $y geq (a-1)$ and $ y leq 1$, which explains the integration limits [you can fill in the blanks].
Finally, since $0 leq X leq 1$ and $0 leq Y leq 1$, you have $0 leq X+Y leq 2$.
answered Nov 26 at 6:32
Aditya Dua
7318
add a comment |
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Answer to your first question: As you yourself noted, $f_Y(a) = 1 ; forall a in [0,1]$, so the $f_Y(a)$ term disappears from the integration. It is not because $int_0^1 f_Y(a)da=1$, which is also true. Also, the limits of integration changed from $(-infty,infty)$ to $[0,1]$ because $f_Y(a)=0$ outside $[0,1]$.
Now you are left with the term $f_X(a-y)$, which equals 1 when $(a-y) in [0,1]$ and 0 otherwise. If $a leq 1$, this holds true when $y leq a$ and $a-y leq 1$. The latter condition can be rewritten as $y geq a-1$. Since $a leq 1$, the condition is equivalent to $y geq 0$. Hence, the limits of integration are $[0,a]$ and over this range $f_x(a-y)=1$, by definition.
By similar reasoning, if $a>1$, $f_X(a-y)=1$ only when $y geq (a-1)$ and $ y leq 1$, which explains the integration limits [you can fill in the blanks].
Finally, since $0 leq X leq 1$ and $0 leq Y leq 1$, you have $0 leq X+Y leq 2$.
answered Nov 26 at 6:32
Aditya Dua
7318
add a comment |
up vote
0
down vote
Answer to your first question: As you yourself noted, $f_Y(a) = 1 ; forall a in [0,1]$, so the $f_Y(a)$ term disappears from the integration. It is not because $int_0^1 f_Y(a)da=1$, which is also true. Also, the limits of integration changed from $(-infty,infty)$ to $[0,1]$ because $f_Y(a)=0$ outside $[0,1]$.
Now you are left with the term $f_X(a-y)$, which equals 1 when $(a-y) in [0,1]$ and 0 otherwise. If $a leq 1$, this holds true when $y leq a$ and $a-y leq 1$. The latter condition can be rewritten as $y geq a-1$. Since $a leq 1$, the condition is equivalent to $y geq 0$. Hence, the limits of integration are $[0,a]$ and over this range $f_x(a-y)=1$, by definition.
By similar reasoning, if $a>1$, $f_X(a-y)=1$ only when $y geq (a-1)$ and $ y leq 1$, which explains the integration limits [you can fill in the blanks].
Finally, since $0 leq X leq 1$ and $0 leq Y leq 1$, you have $0 leq X+Y leq 2$.
answered Nov 26 at 6:32
Aditya Dua
7318
add a comment |
up vote
0
down vote
up vote
0
down vote
Answer to your first question: As you yourself noted, $f_Y(a) = 1 ; forall a in [0,1]$, so the $f_Y(a)$ term disappears from the integration. It is not because $int_0^1 f_Y(a)da=1$, which is also true. Also, the limits of integration changed from $(-infty,infty)$ to $[0,1]$ because $f_Y(a)=0$ outside $[0,1]$.
Now you are left with the term $f_X(a-y)$, which equals 1 when $(a-y) in [0,1]$ and 0 otherwise. If $a leq 1$, this holds true when $y leq a$ and $a-y leq 1$. The latter condition can be rewritten as $y geq a-1$. Since $a leq 1$, the condition is equivalent to $y geq 0$. Hence, the limits of integration are $[0,a]$ and over this range $f_x(a-y)=1$, by definition.
By similar reasoning, if $a>1$, $f_X(a-y)=1$ only when $y geq (a-1)$ and $ y leq 1$, which explains the integration limits [you can fill in the blanks].
Finally, since $0 leq X leq 1$ and $0 leq Y leq 1$, you have $0 leq X+Y leq 2$.
answered Nov 26 at 6:32
Aditya Dua
7318
Answer to your first question: As you yourself noted, $f_Y(a) = 1 ; forall a in [0,1]$, so the $f_Y(a)$ term disappears from the integration. It is not because $int_0^1 f_Y(a)da=1$, which is also true. Also, the limits of integration changed from $(-infty,infty)$ to $[0,1]$ because $f_Y(a)=0$ outside $[0,1]$.
Now you are left with the term $f_X(a-y)$, which equals 1 when $(a-y) in [0,1]$ and 0 otherwise. If $a leq 1$, this holds true when $y leq a$ and $a-y leq 1$. The latter condition can be rewritten as $y geq a-1$. Since $a leq 1$, the condition is equivalent to $y geq 0$. Hence, the limits of integration are $[0,a]$ and over this range $f_x(a-y)=1$, by definition.
By similar reasoning, if $a>1$, $f_X(a-y)=1$ only when $y geq (a-1)$ and $ y leq 1$, which explains the integration limits [you can fill in the blanks].
Finally, since $0 leq X leq 1$ and $0 leq Y leq 1$, you have $0 leq X+Y leq 2$.
answered Nov 26 at 6:32
Aditya Dua
7318
answered Nov 26 at 6:32
Aditya Dua
7318
answered Nov 26 at 6:32
Aditya Dua
7318
answered Nov 26 at 6:32
Aditya Dua
7318
7318
add a comment |
add a comment |
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The density of $f_Y$ is $1$ so there's no need to write it. It should be clear that adding two $(0, 1)$ variables means the sum can be at most $2.$ Also, uniform plus uniform is triangle.
– Sean Roberson
Nov 26 at 6:12
how'd they know to split it into the case where $0 leq a leq 1$ and $1 < a < 2$? why not like $0 leq a leq 1/2$ etc?
– stackofhay42
Nov 26 at 6:13