Functions of uniform density












0












$begingroup$


Consider the random variable X with the uniform
density having $α = 1$ and $β = 3$.
(a) Use the result of Example 2 to find the probability
density of $Y = |X|$



In example two they showed that such a function(Y=|X|) would have the density of the form:
$g(y) = begin{cases} f(y) + f(-y) space text{for} space y>0 \ 0 space text{elsewhere} end{cases}$



I have learnt two techinques, distribution function technique and transformation in one variable technique.
The answer given in the book:
$frac{1}{8} y^{−3/4} $ for $ 0 < y < 1$ and
$g(y) =frac{1}{4}$ for $1 <y < 3$



I'm totally lost in this question



Shouldn't there be distribution only for $1<y<3$? Why is $0<y<1$ there?
Even if both are there I'm not getting anything close to the two given answers, just getting weird answers like 1 and 1/2






random-variables







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you think how the distribution of $X$ looks like? And what is the result of example 2?
    $endgroup$
    – callculus
    Dec 3 '18 at 11:00












  • $begingroup$
    Apologies, have now put in info about example 2
    $endgroup$
    – Sumukh Sai
    Dec 3 '18 at 11:14










  • $begingroup$
    Looks to me the distribution of $|X|$ is the same as that of $X$. Have you quoted the question correctly?
    $endgroup$
    – StubbornAtom
    Dec 3 '18 at 12:24










  • $begingroup$
    The question is quoted correctly
    $endgroup$
    – Sumukh Sai
    Dec 3 '18 at 12:32
















0












$begingroup$


Consider the random variable X with the uniform
density having $α = 1$ and $β = 3$.
(a) Use the result of Example 2 to find the probability
density of $Y = |X|$



In example two they showed that such a function(Y=|X|) would have the density of the form:
$g(y) = begin{cases} f(y) + f(-y) space text{for} space y>0 \ 0 space text{elsewhere} end{cases}$



I have learnt two techinques, distribution function technique and transformation in one variable technique.
The answer given in the book:
$frac{1}{8} y^{−3/4} $ for $ 0 < y < 1$ and
$g(y) =frac{1}{4}$ for $1 <y < 3$



I'm totally lost in this question



Shouldn't there be distribution only for $1<y<3$? Why is $0<y<1$ there?
Even if both are there I'm not getting anything close to the two given answers, just getting weird answers like 1 and 1/2






random-variables







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you think how the distribution of $X$ looks like? And what is the result of example 2?
    $endgroup$
    – callculus
    Dec 3 '18 at 11:00












  • $begingroup$
    Apologies, have now put in info about example 2
    $endgroup$
    – Sumukh Sai
    Dec 3 '18 at 11:14










  • $begingroup$
    Looks to me the distribution of $|X|$ is the same as that of $X$. Have you quoted the question correctly?
    $endgroup$
    – StubbornAtom
    Dec 3 '18 at 12:24










  • $begingroup$
    The question is quoted correctly
    $endgroup$
    – Sumukh Sai
    Dec 3 '18 at 12:32














0












0








0


0



$begingroup$


Consider the random variable X with the uniform
density having $α = 1$ and $β = 3$.
(a) Use the result of Example 2 to find the probability
density of $Y = |X|$



In example two they showed that such a function(Y=|X|) would have the density of the form:
$g(y) = begin{cases} f(y) + f(-y) space text{for} space y>0 \ 0 space text{elsewhere} end{cases}$



I have learnt two techinques, distribution function technique and transformation in one variable technique.
The answer given in the book:
$frac{1}{8} y^{−3/4} $ for $ 0 < y < 1$ and
$g(y) =frac{1}{4}$ for $1 <y < 3$



I'm totally lost in this question



Shouldn't there be distribution only for $1<y<3$? Why is $0<y<1$ there?
Even if both are there I'm not getting anything close to the two given answers, just getting weird answers like 1 and 1/2






random-variables







share|cite|improve this question











$endgroup$




Consider the random variable X with the uniform
density having $α = 1$ and $β = 3$.
(a) Use the result of Example 2 to find the probability
density of $Y = |X|$



In example two they showed that such a function(Y=|X|) would have the density of the form:
$g(y) = begin{cases} f(y) + f(-y) space text{for} space y>0 \ 0 space text{elsewhere} end{cases}$



I have learnt two techinques, distribution function technique and transformation in one variable technique.
The answer given in the book:
$frac{1}{8} y^{−3/4} $ for $ 0 < y < 1$ and
$g(y) =frac{1}{4}$ for $1 <y < 3$



I'm totally lost in this question



Shouldn't there be distribution only for $1<y<3$? Why is $0<y<1$ there?
Even if both are there I'm not getting anything close to the two given answers, just getting weird answers like 1 and 1/2






random-variables




random-variables






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 11:28







Sumukh Sai

















asked Dec 3 '18 at 10:54









Sumukh SaiSumukh Sai

206




206












  • $begingroup$
    What do you think how the distribution of $X$ looks like? And what is the result of example 2?
    $endgroup$
    – callculus
    Dec 3 '18 at 11:00












  • $begingroup$
    Apologies, have now put in info about example 2
    $endgroup$
    – Sumukh Sai
    Dec 3 '18 at 11:14










  • $begingroup$
    Looks to me the distribution of $|X|$ is the same as that of $X$. Have you quoted the question correctly?
    $endgroup$
    – StubbornAtom
    Dec 3 '18 at 12:24










  • $begingroup$
    The question is quoted correctly
    $endgroup$
    – Sumukh Sai
    Dec 3 '18 at 12:32


















  • $begingroup$
    What do you think how the distribution of $X$ looks like? And what is the result of example 2?
    $endgroup$
    – callculus
    Dec 3 '18 at 11:00












  • $begingroup$
    Apologies, have now put in info about example 2
    $endgroup$
    – Sumukh Sai
    Dec 3 '18 at 11:14










  • $begingroup$
    Looks to me the distribution of $|X|$ is the same as that of $X$. Have you quoted the question correctly?
    $endgroup$
    – StubbornAtom
    Dec 3 '18 at 12:24










  • $begingroup$
    The question is quoted correctly
    $endgroup$
    – Sumukh Sai
    Dec 3 '18 at 12:32
















$begingroup$
What do you think how the distribution of $X$ looks like? And what is the result of example 2?
$endgroup$
– callculus
Dec 3 '18 at 11:00






$begingroup$
What do you think how the distribution of $X$ looks like? And what is the result of example 2?
$endgroup$
– callculus
Dec 3 '18 at 11:00














$begingroup$
Apologies, have now put in info about example 2
$endgroup$
– Sumukh Sai
Dec 3 '18 at 11:14




$begingroup$
Apologies, have now put in info about example 2
$endgroup$
– Sumukh Sai
Dec 3 '18 at 11:14












$begingroup$
Looks to me the distribution of $|X|$ is the same as that of $X$. Have you quoted the question correctly?
$endgroup$
– StubbornAtom
Dec 3 '18 at 12:24




$begingroup$
Looks to me the distribution of $|X|$ is the same as that of $X$. Have you quoted the question correctly?
$endgroup$
– StubbornAtom
Dec 3 '18 at 12:24












$begingroup$
The question is quoted correctly
$endgroup$
– Sumukh Sai
Dec 3 '18 at 12:32




$begingroup$
The question is quoted correctly
$endgroup$
– Sumukh Sai
Dec 3 '18 at 12:32










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