Let $U_1$ and $U_2$ be independent and uniform on $[0, 1]$, find and sketch the density function of $S = U_1...












0














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







share|cite|improve this question




















  • 2




    what difficulty did you face when you tried the formula?
    – Siong Thye Goh
    Nov 30 '18 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 '18 at 1:58
















0














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







share|cite|improve this question




















  • 2




    what difficulty did you face when you tried the formula?
    – Siong Thye Goh
    Nov 30 '18 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 '18 at 1:58














0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 2:17









Tianlalu

3,09621038




3,09621038










asked Nov 30 '18 at 1:43









pecopeco

758




758








  • 2




    what difficulty did you face when you tried the formula?
    – Siong Thye Goh
    Nov 30 '18 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 '18 at 1:58














  • 2




    what difficulty did you face when you tried the formula?
    – Siong Thye Goh
    Nov 30 '18 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 '18 at 1:58








2




2




what difficulty did you face when you tried the formula?
– Siong Thye Goh
Nov 30 '18 at 1:45




what difficulty did you face when you tried the formula?
– Siong Thye Goh
Nov 30 '18 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 '18 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 '18 at 1:58










2 Answers
2






active

oldest

votes


















0














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
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$






share|cite|improve this answer





















  • 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 '18 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 '18 at 3:05



















0














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 $t$.






share|cite|improve this answer























  • 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 '18 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 '18 at 2:22











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019506%2flet-u-1-and-u-2-be-independent-and-uniform-on-0-1-find-and-sketch-the%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














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
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$






share|cite|improve this answer





















  • 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 '18 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 '18 at 3:05
















0














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
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$






share|cite|improve this answer





















  • 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 '18 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 '18 at 3:05














0












0








0






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
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$






share|cite|improve this answer












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
$$f_S(z) = int_{z-1}^1 , mathrm{d}x = 2-z.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 '18 at 2:15









Timothy HedgeworthTimothy 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 3:05











0














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 $t$.






share|cite|improve this answer























  • 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 '18 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 '18 at 2:22
















0














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 $t$.






share|cite|improve this answer























  • 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 '18 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 '18 at 2:22














0












0








0






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 $t$.






share|cite|improve this answer














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 $t$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 2:12

























answered Nov 30 '18 at 2:05









Siong Thye GohSiong Thye Goh

99.6k1464117




99.6k1464117












  • 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 2:22


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019506%2flet-u-1-and-u-2-be-independent-and-uniform-on-0-1-find-and-sketch-the%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more