Let $U_1$ and $U_2$ be independent and uniform on $[0, 1]$, find and sketch the density function of $S = U_1...
I am stuck on this problem during my review for my stats test.
I know I have to use the convolution formula, and I understand that:
$f_{U_1}(U_1) = 1$ for $0≤U_1≤1$
$f_{U_2}(U_2) = 1$ for $0≤U_2≤1$
but I do not know how to continue on from there. How do I use the convolution formula in this question? Thanks
statistics
edited Nov 30 at 2:17
Tianlalu
3,12321038
asked Nov 30 at 1:43
peco
758
add a comment |
I am stuck on this problem during my review for my stats test.
I know I have to use the convolution formula, and I understand that:
$f_{U_1}(U_1) = 1$ for $0≤U_1≤1$
$f_{U_2}(U_2) = 1$ for $0≤U_2≤1$
but I do not know how to continue on from there. How do I use the convolution formula in this question? Thanks
statistics
edited Nov 30 at 2:17
Tianlalu
3,12321038
asked Nov 30 at 1:43
peco
758
2
what difficulty did you face when you tried the formula?
– Siong Thye Goh
Nov 30 at 1:45
I'm not sure how to even set up the equation. I'm guessing I start off with $int dU_1dU_2$. Is that correct?
– peco
Nov 30 at 1:58
add a comment |
I am stuck on this problem during my review for my stats test.
I know I have to use the convolution formula, and I understand that:
$f_{U_1}(U_1) = 1$ for $0≤U_1≤1$
$f_{U_2}(U_2) = 1$ for $0≤U_2≤1$
but I do not know how to continue on from there. How do I use the convolution formula in this question? Thanks
statistics
edited Nov 30 at 2:17
Tianlalu
3,12321038
asked Nov 30 at 1:43
peco
758
I am stuck on this problem during my review for my stats test.
I know I have to use the convolution formula, and I understand that:
$f_{U_1}(U_1) = 1$ for $0≤U_1≤1$
$f_{U_2}(U_2) = 1$ for $0≤U_2≤1$
but I do not know how to continue on from there. How do I use the convolution formula in this question? Thanks
statistics
statistics
edited Nov 30 at 2:17
Tianlalu
3,12321038
asked Nov 30 at 1:43
peco
758
edited Nov 30 at 2:17
Tianlalu
3,12321038
asked Nov 30 at 1:43
peco
758
edited Nov 30 at 2:17
Tianlalu
3,12321038
edited Nov 30 at 2:17
Tianlalu
3,12321038
edited Nov 30 at 2:17
Tianlalu
3,12321038
3,12321038
asked Nov 30 at 1:43
peco
758
asked Nov 30 at 1:43
peco
758
asked Nov 30 at 1:43
peco
758
758
2
what difficulty did you face when you tried the formula?
– Siong Thye Goh
Nov 30 at 1:45
I'm not sure how to even set up the equation. I'm guessing I start off with $int dU_1dU_2$. Is that correct?
– peco
Nov 30 at 1:58
add a comment |
2
what difficulty did you face when you tried the formula?
– Siong Thye Goh
Nov 30 at 1:45
I'm not sure how to even set up the equation. I'm guessing I start off with $int dU_1dU_2$. Is that correct?
– peco
Nov 30 at 1:58
2
2
what difficulty did you face when you tried the formula?
– Siong Thye Goh
Nov 30 at 1:45
what difficulty did you face when you tried the formula?
– Siong Thye Goh
Nov 30 at 1:45
I'm not sure how to even set up the equation. I'm guessing I start off with $int dU_1dU_2$. Is that correct?
– peco
Nov 30 at 1:58
I'm not sure how to even set up the equation. I'm guessing I start off with $int dU_1dU_2$. Is that correct?
– peco
Nov 30 at 1:58
add a comment |
2 Answers
2
active
oldest
votes
If you are OK with the convolution formula then it is just a bit of computation. Letting $f_{U_1}(x) = f_{U_2}(x) = 1$ for $x in [0,1]$ and $0$ elsewhere, and $S = U_1 + U_2$. The convolution of $f_{U_1}$ and $f_{U_2}$ gives us the probability density function of $f_S$, it is
$$ f_S(z) = int_{-infty}^infty f_{U_1}(z-x)f_{U_2}(x) , mathrm{d}x. $$
Since $f_{U_2}(x) = 1$ on $[0,1]$ we have
$$f_S(z) = int_0^1 f_{U_1}(z-x) , mathrm{d}x.$$
We now have to consider cases since $f_{U_1}(z-x) = 1$ only when $ 0leq z-x leq 1$, or equivalently when $z-1leq xleq z$. Thus, for $0leq zleq 1$ we have
$$f_S(z) = int_0^z , mathrm{d}x = z.$$
And, for $1leq zleq 2$, we have
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$
answered Nov 30 at 2:15
Timothy Hedgeworth
1366
thanks, but I don't understand why we need to consider $1≤z≤2$. I have also seen in the textbook answer that we need to have 2 regions, but is it because we have to consider the region of S which goes from 0 to 2 as $S = U_1 + U_2$?
– peco
Nov 30 at 2:21
1
We need to consider $1leq zleq 2$ since it may be the case that $0leq z-x leq 1$ and so $f_{U_1}(z-x) = 1.$ As mentioned above, $f_{U_1}(z-x) = 1$ is satisfied whenever $z-1leq xleq z$ and note that we are only considering $0leq xleq 1$ from the limits of the integral. We need to consider the two regions since both inequalities can be satisfied for each region: for $0leq zleq 1$ - the inequalities are satisfied whenever $0leq xleq z$; for $1leq z leq 2$ - the inequalities are satisfied whenever $z-1leq xleq 1$. Hopefully this helps?
– Timothy Hedgeworth
Nov 30 at 3:05
add a comment |
Let $s in [0,2]$
begin{align}
f_S(s) &= int_{-infty}^infty f_{U_1}(t)f_{U_2}(s-t) , dt \
&= int_0^1 1cdot f_{U_2}(s-t) , dt
end{align}
Now we just need to integrate over the region when $f_{U_2}$ is positive.
$$0 le s-t le 1$$
$$-1 le t-s le 0$$
$$s-1 le t le s$$
Hence
$$f_S(s) = int_{max(0,s-1)}^{min{(1,s)}} 1 , dt$$
I will leave the simplication as an exercise.
You might like to consider when $s ge 1$ and $s<1$.
Remark: In the convoltuion formula, we are integrating only over one variable $t$.
answered Nov 30 at 2:05
Siong Thye Goh
98.9k1464116
sorry for asking so much, but I know min(1,s) means the minimum. But how do you find what it equals to?
– peco
Nov 30 at 2:16
If $s< 1$, then we have $min(1,s)=s$ and if we have $s ge 1$, then we have $min(1,s)=1$.
– Siong Thye Goh
Nov 30 at 2:22
add a comment |
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2 Answers
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2 Answers
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If you are OK with the convolution formula then it is just a bit of computation. Letting $f_{U_1}(x) = f_{U_2}(x) = 1$ for $x in [0,1]$ and $0$ elsewhere, and $S = U_1 + U_2$. The convolution of $f_{U_1}$ and $f_{U_2}$ gives us the probability density function of $f_S$, it is
$$ f_S(z) = int_{-infty}^infty f_{U_1}(z-x)f_{U_2}(x) , mathrm{d}x. $$
Since $f_{U_2}(x) = 1$ on $[0,1]$ we have
$$f_S(z) = int_0^1 f_{U_1}(z-x) , mathrm{d}x.$$
We now have to consider cases since $f_{U_1}(z-x) = 1$ only when $ 0leq z-x leq 1$, or equivalently when $z-1leq xleq z$. Thus, for $0leq zleq 1$ we have
$$f_S(z) = int_0^z , mathrm{d}x = z.$$
And, for $1leq zleq 2$, we have
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$
answered Nov 30 at 2:15
Timothy Hedgeworth
1366
thanks, but I don't understand why we need to consider $1≤z≤2$. I have also seen in the textbook answer that we need to have 2 regions, but is it because we have to consider the region of S which goes from 0 to 2 as $S = U_1 + U_2$?
– peco
Nov 30 at 2:21
1
We need to consider $1leq zleq 2$ since it may be the case that $0leq z-x leq 1$ and so $f_{U_1}(z-x) = 1.$ As mentioned above, $f_{U_1}(z-x) = 1$ is satisfied whenever $z-1leq xleq z$ and note that we are only considering $0leq xleq 1$ from the limits of the integral. We need to consider the two regions since both inequalities can be satisfied for each region: for $0leq zleq 1$ - the inequalities are satisfied whenever $0leq xleq z$; for $1leq z leq 2$ - the inequalities are satisfied whenever $z-1leq xleq 1$. Hopefully this helps?
– Timothy Hedgeworth
Nov 30 at 3:05
add a comment |
If you are OK with the convolution formula then it is just a bit of computation. Letting $f_{U_1}(x) = f_{U_2}(x) = 1$ for $x in [0,1]$ and $0$ elsewhere, and $S = U_1 + U_2$. The convolution of $f_{U_1}$ and $f_{U_2}$ gives us the probability density function of $f_S$, it is
$$ f_S(z) = int_{-infty}^infty f_{U_1}(z-x)f_{U_2}(x) , mathrm{d}x. $$
Since $f_{U_2}(x) = 1$ on $[0,1]$ we have
$$f_S(z) = int_0^1 f_{U_1}(z-x) , mathrm{d}x.$$
We now have to consider cases since $f_{U_1}(z-x) = 1$ only when $ 0leq z-x leq 1$, or equivalently when $z-1leq xleq z$. Thus, for $0leq zleq 1$ we have
$$f_S(z) = int_0^z , mathrm{d}x = z.$$
And, for $1leq zleq 2$, we have
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$
answered Nov 30 at 2:15
Timothy Hedgeworth
1366
thanks, but I don't understand why we need to consider $1≤z≤2$. I have also seen in the textbook answer that we need to have 2 regions, but is it because we have to consider the region of S which goes from 0 to 2 as $S = U_1 + U_2$?
– peco
Nov 30 at 2:21
1
We need to consider $1leq zleq 2$ since it may be the case that $0leq z-x leq 1$ and so $f_{U_1}(z-x) = 1.$ As mentioned above, $f_{U_1}(z-x) = 1$ is satisfied whenever $z-1leq xleq z$ and note that we are only considering $0leq xleq 1$ from the limits of the integral. We need to consider the two regions since both inequalities can be satisfied for each region: for $0leq zleq 1$ - the inequalities are satisfied whenever $0leq xleq z$; for $1leq z leq 2$ - the inequalities are satisfied whenever $z-1leq xleq 1$. Hopefully this helps?
– Timothy Hedgeworth
Nov 30 at 3:05
add a comment |
If you are OK with the convolution formula then it is just a bit of computation. Letting $f_{U_1}(x) = f_{U_2}(x) = 1$ for $x in [0,1]$ and $0$ elsewhere, and $S = U_1 + U_2$. The convolution of $f_{U_1}$ and $f_{U_2}$ gives us the probability density function of $f_S$, it is
$$ f_S(z) = int_{-infty}^infty f_{U_1}(z-x)f_{U_2}(x) , mathrm{d}x. $$
Since $f_{U_2}(x) = 1$ on $[0,1]$ we have
$$f_S(z) = int_0^1 f_{U_1}(z-x) , mathrm{d}x.$$
We now have to consider cases since $f_{U_1}(z-x) = 1$ only when $ 0leq z-x leq 1$, or equivalently when $z-1leq xleq z$. Thus, for $0leq zleq 1$ we have
$$f_S(z) = int_0^z , mathrm{d}x = z.$$
And, for $1leq zleq 2$, we have
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$
answered Nov 30 at 2:15
Timothy Hedgeworth
1366
If you are OK with the convolution formula then it is just a bit of computation. Letting $f_{U_1}(x) = f_{U_2}(x) = 1$ for $x in [0,1]$ and $0$ elsewhere, and $S = U_1 + U_2$. The convolution of $f_{U_1}$ and $f_{U_2}$ gives us the probability density function of $f_S$, it is
$$ f_S(z) = int_{-infty}^infty f_{U_1}(z-x)f_{U_2}(x) , mathrm{d}x. $$
Since $f_{U_2}(x) = 1$ on $[0,1]$ we have
$$f_S(z) = int_0^1 f_{U_1}(z-x) , mathrm{d}x.$$
We now have to consider cases since $f_{U_1}(z-x) = 1$ only when $ 0leq z-x leq 1$, or equivalently when $z-1leq xleq z$. Thus, for $0leq zleq 1$ we have
$$f_S(z) = int_0^z , mathrm{d}x = z.$$
And, for $1leq zleq 2$, we have
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$
answered Nov 30 at 2:15
Timothy Hedgeworth
1366
answered Nov 30 at 2:15
Timothy Hedgeworth
1366
answered Nov 30 at 2:15
Timothy Hedgeworth
1366
answered Nov 30 at 2:15
Timothy Hedgeworth
1366
1366
thanks, but I don't understand why we need to consider $1≤z≤2$. I have also seen in the textbook answer that we need to have 2 regions, but is it because we have to consider the region of S which goes from 0 to 2 as $S = U_1 + U_2$?
– peco
Nov 30 at 2:21
1
We need to consider $1leq zleq 2$ since it may be the case that $0leq z-x leq 1$ and so $f_{U_1}(z-x) = 1.$ As mentioned above, $f_{U_1}(z-x) = 1$ is satisfied whenever $z-1leq xleq z$ and note that we are only considering $0leq xleq 1$ from the limits of the integral. We need to consider the two regions since both inequalities can be satisfied for each region: for $0leq zleq 1$ - the inequalities are satisfied whenever $0leq xleq z$; for $1leq z leq 2$ - the inequalities are satisfied whenever $z-1leq xleq 1$. Hopefully this helps?
– Timothy Hedgeworth
Nov 30 at 3:05
add a comment |
thanks, but I don't understand why we need to consider $1≤z≤2$. I have also seen in the textbook answer that we need to have 2 regions, but is it because we have to consider the region of S which goes from 0 to 2 as $S = U_1 + U_2$?
– peco
Nov 30 at 2:21
1
We need to consider $1leq zleq 2$ since it may be the case that $0leq z-x leq 1$ and so $f_{U_1}(z-x) = 1.$ As mentioned above, $f_{U_1}(z-x) = 1$ is satisfied whenever $z-1leq xleq z$ and note that we are only considering $0leq xleq 1$ from the limits of the integral. We need to consider the two regions since both inequalities can be satisfied for each region: for $0leq zleq 1$ - the inequalities are satisfied whenever $0leq xleq z$; for $1leq z leq 2$ - the inequalities are satisfied whenever $z-1leq xleq 1$. Hopefully this helps?
– Timothy Hedgeworth
Nov 30 at 3:05
thanks, but I don't understand why we need to consider $1≤z≤2$. I have also seen in the textbook answer that we need to have 2 regions, but is it because we have to consider the region of S which goes from 0 to 2 as $S = U_1 + U_2$?
– peco
Nov 30 at 2:21
thanks, but I don't understand why we need to consider $1≤z≤2$. I have also seen in the textbook answer that we need to have 2 regions, but is it because we have to consider the region of S which goes from 0 to 2 as $S = U_1 + U_2$?
– peco
Nov 30 at 2:21
1
1
We need to consider $1leq zleq 2$ since it may be the case that $0leq z-x leq 1$ and so $f_{U_1}(z-x) = 1.$ As mentioned above, $f_{U_1}(z-x) = 1$ is satisfied whenever $z-1leq xleq z$ and note that we are only considering $0leq xleq 1$ from the limits of the integral. We need to consider the two regions since both inequalities can be satisfied for each region: for $0leq zleq 1$ - the inequalities are satisfied whenever $0leq xleq z$; for $1leq z leq 2$ - the inequalities are satisfied whenever $z-1leq xleq 1$. Hopefully this helps?
– Timothy Hedgeworth
Nov 30 at 3:05
We need to consider $1leq zleq 2$ since it may be the case that $0leq z-x leq 1$ and so $f_{U_1}(z-x) = 1.$ As mentioned above, $f_{U_1}(z-x) = 1$ is satisfied whenever $z-1leq xleq z$ and note that we are only considering $0leq xleq 1$ from the limits of the integral. We need to consider the two regions since both inequalities can be satisfied for each region: for $0leq zleq 1$ - the inequalities are satisfied whenever $0leq xleq z$; for $1leq z leq 2$ - the inequalities are satisfied whenever $z-1leq xleq 1$. Hopefully this helps?
– Timothy Hedgeworth
Nov 30 at 3:05
add a comment |
Let $s in [0,2]$
begin{align}
f_S(s) &= int_{-infty}^infty f_{U_1}(t)f_{U_2}(s-t) , dt \
&= int_0^1 1cdot f_{U_2}(s-t) , dt
end{align}
Now we just need to integrate over the region when $f_{U_2}$ is positive.
$$0 le s-t le 1$$
$$-1 le t-s le 0$$
$$s-1 le t le s$$
Hence
$$f_S(s) = int_{max(0,s-1)}^{min{(1,s)}} 1 , dt$$
I will leave the simplication as an exercise.
You might like to consider when $s ge 1$ and $s<1$.
Remark: In the convoltuion formula, we are integrating only over one variable $t$.
answered Nov 30 at 2:05
Siong Thye Goh
98.9k1464116
sorry for asking so much, but I know min(1,s) means the minimum. But how do you find what it equals to?
– peco
Nov 30 at 2:16
If $s< 1$, then we have $min(1,s)=s$ and if we have $s ge 1$, then we have $min(1,s)=1$.
– Siong Thye Goh
Nov 30 at 2:22
add a comment |
Let $s in [0,2]$
begin{align}
f_S(s) &= int_{-infty}^infty f_{U_1}(t)f_{U_2}(s-t) , dt \
&= int_0^1 1cdot f_{U_2}(s-t) , dt
end{align}
Now we just need to integrate over the region when $f_{U_2}$ is positive.
$$0 le s-t le 1$$
$$-1 le t-s le 0$$
$$s-1 le t le s$$
Hence
$$f_S(s) = int_{max(0,s-1)}^{min{(1,s)}} 1 , dt$$
I will leave the simplication as an exercise.
You might like to consider when $s ge 1$ and $s<1$.
Remark: In the convoltuion formula, we are integrating only over one variable $t$.
answered Nov 30 at 2:05
Siong Thye Goh
98.9k1464116
sorry for asking so much, but I know min(1,s) means the minimum. But how do you find what it equals to?
– peco
Nov 30 at 2:16
If $s< 1$, then we have $min(1,s)=s$ and if we have $s ge 1$, then we have $min(1,s)=1$.
– Siong Thye Goh
Nov 30 at 2:22
add a comment |
Let $s in [0,2]$
begin{align}
f_S(s) &= int_{-infty}^infty f_{U_1}(t)f_{U_2}(s-t) , dt \
&= int_0^1 1cdot f_{U_2}(s-t) , dt
end{align}
Now we just need to integrate over the region when $f_{U_2}$ is positive.
$$0 le s-t le 1$$
$$-1 le t-s le 0$$
$$s-1 le t le s$$
Hence
$$f_S(s) = int_{max(0,s-1)}^{min{(1,s)}} 1 , dt$$
I will leave the simplication as an exercise.
You might like to consider when $s ge 1$ and $s<1$.
Remark: In the convoltuion formula, we are integrating only over one variable $t$.
answered Nov 30 at 2:05
Siong Thye Goh
98.9k1464116
Let $s in [0,2]$
begin{align}
f_S(s) &= int_{-infty}^infty f_{U_1}(t)f_{U_2}(s-t) , dt \
&= int_0^1 1cdot f_{U_2}(s-t) , dt
end{align}
Now we just need to integrate over the region when $f_{U_2}$ is positive.
$$0 le s-t le 1$$
$$-1 le t-s le 0$$
$$s-1 le t le s$$
Hence
$$f_S(s) = int_{max(0,s-1)}^{min{(1,s)}} 1 , dt$$
I will leave the simplication as an exercise.
You might like to consider when $s ge 1$ and $s<1$.
Remark: In the convoltuion formula, we are integrating only over one variable $t$.
answered Nov 30 at 2:05
Siong Thye Goh
98.9k1464116
edited Nov 30 at 2:12
answered Nov 30 at 2:05
Siong Thye Goh
98.9k1464116
answered Nov 30 at 2:05
Siong Thye Goh
98.9k1464116
answered Nov 30 at 2:05
Siong Thye Goh
98.9k1464116
98.9k1464116
sorry for asking so much, but I know min(1,s) means the minimum. But how do you find what it equals to?
– peco
Nov 30 at 2:16
If $s< 1$, then we have $min(1,s)=s$ and if we have $s ge 1$, then we have $min(1,s)=1$.
– Siong Thye Goh
Nov 30 at 2:22
add a comment |
sorry for asking so much, but I know min(1,s) means the minimum. But how do you find what it equals to?
– peco
Nov 30 at 2:16
If $s< 1$, then we have $min(1,s)=s$ and if we have $s ge 1$, then we have $min(1,s)=1$.
– Siong Thye Goh
Nov 30 at 2:22
sorry for asking so much, but I know min(1,s) means the minimum. But how do you find what it equals to?
– peco
Nov 30 at 2:16
sorry for asking so much, but I know min(1,s) means the minimum. But how do you find what it equals to?
– peco
Nov 30 at 2:16
If $s< 1$, then we have $min(1,s)=s$ and if we have $s ge 1$, then we have $min(1,s)=1$.
– Siong Thye Goh
Nov 30 at 2:22
If $s< 1$, then we have $min(1,s)=s$ and if we have $s ge 1$, then we have $min(1,s)=1$.
– Siong Thye Goh
Nov 30 at 2:22
add a comment |
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what difficulty did you face when you tried the formula?
– Siong Thye Goh
Nov 30 at 1:45
I'm not sure how to even set up the equation. I'm guessing I start off with $int dU_1dU_2$. Is that correct?
– peco
Nov 30 at 1:58