How to determine cumulative distribution function of squared random variable?












1












$begingroup$


I don't understand how can I get the



$F(x)$(cumulative distribution funciton)



given only



$X = RND^2$



where RND means (continuous) random variable, the paper tells the answer is $F(x) = sqrt x$
, if x $epsilon (0,1)$ it seems that I need to take square root but why?



I know $F(x) = p(X < x)$ but from here I can't proceed.



enter image description here






probability probability-theory probability-distributions random-variables







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know the distribution function for RND?
    $endgroup$
    – herb steinberg
    Dec 13 '18 at 17:30










  • $begingroup$
    I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
    $endgroup$
    – Levente Bartos
    Dec 13 '18 at 17:40










  • $begingroup$
    I assume it is uniformly distributed
    $endgroup$
    – Levente Bartos
    Dec 13 '18 at 17:42
















1












$begingroup$


I don't understand how can I get the



$F(x)$(cumulative distribution funciton)



given only



$X = RND^2$



where RND means (continuous) random variable, the paper tells the answer is $F(x) = sqrt x$
, if x $epsilon (0,1)$ it seems that I need to take square root but why?



I know $F(x) = p(X < x)$ but from here I can't proceed.



enter image description here






probability probability-theory probability-distributions random-variables







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know the distribution function for RND?
    $endgroup$
    – herb steinberg
    Dec 13 '18 at 17:30










  • $begingroup$
    I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
    $endgroup$
    – Levente Bartos
    Dec 13 '18 at 17:40










  • $begingroup$
    I assume it is uniformly distributed
    $endgroup$
    – Levente Bartos
    Dec 13 '18 at 17:42














1












1








1





$begingroup$


I don't understand how can I get the



$F(x)$(cumulative distribution funciton)



given only



$X = RND^2$



where RND means (continuous) random variable, the paper tells the answer is $F(x) = sqrt x$
, if x $epsilon (0,1)$ it seems that I need to take square root but why?



I know $F(x) = p(X < x)$ but from here I can't proceed.



enter image description here






probability probability-theory probability-distributions random-variables







share|cite|improve this question











$endgroup$




I don't understand how can I get the



$F(x)$(cumulative distribution funciton)



given only



$X = RND^2$



where RND means (continuous) random variable, the paper tells the answer is $F(x) = sqrt x$
, if x $epsilon (0,1)$ it seems that I need to take square root but why?



I know $F(x) = p(X < x)$ but from here I can't proceed.



enter image description here






probability probability-theory probability-distributions random-variables




probability probability-theory probability-distributions random-variables






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 17:39







Levente Bartos

















asked Dec 13 '18 at 17:26









Levente BartosLevente Bartos

186




186












  • $begingroup$
    Do you know the distribution function for RND?
    $endgroup$
    – herb steinberg
    Dec 13 '18 at 17:30










  • $begingroup$
    I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
    $endgroup$
    – Levente Bartos
    Dec 13 '18 at 17:40










  • $begingroup$
    I assume it is uniformly distributed
    $endgroup$
    – Levente Bartos
    Dec 13 '18 at 17:42


















  • $begingroup$
    Do you know the distribution function for RND?
    $endgroup$
    – herb steinberg
    Dec 13 '18 at 17:30










  • $begingroup$
    I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
    $endgroup$
    – Levente Bartos
    Dec 13 '18 at 17:40










  • $begingroup$
    I assume it is uniformly distributed
    $endgroup$
    – Levente Bartos
    Dec 13 '18 at 17:42
















$begingroup$
Do you know the distribution function for RND?
$endgroup$
– herb steinberg
Dec 13 '18 at 17:30




$begingroup$
Do you know the distribution function for RND?
$endgroup$
– herb steinberg
Dec 13 '18 at 17:30












$begingroup$
I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:40




$begingroup$
I have added a picture about the problem, i assume if I get this kind of question during an exam they will give only the first line X = RND^2 and then i should continue?
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:40












$begingroup$
I assume it is uniformly distributed
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:42




$begingroup$
I assume it is uniformly distributed
$endgroup$
– Levente Bartos
Dec 13 '18 at 17:42










1 Answer
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$begingroup$

You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.



$P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$



Now you plug in the value $sqrt x$ into the cdf of $Y$:



$$F_Y(y)=begin{cases}
0 & text{for } y < 0 \
y & text{for } y in [0,1) \
1 & text{for } x ge 1
end{cases}$$



To obtain the pdf you just differentiate the cdf of Y.






share|cite|improve this answer









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    1 Answer
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    $begingroup$

    You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.



    $P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$



    Now you plug in the value $sqrt x$ into the cdf of $Y$:



    $$F_Y(y)=begin{cases}
    0 & text{for } y < 0 \
    y & text{for } y in [0,1) \
    1 & text{for } x ge 1
    end{cases}$$



    To obtain the pdf you just differentiate the cdf of Y.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.



      $P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$



      Now you plug in the value $sqrt x$ into the cdf of $Y$:



      $$F_Y(y)=begin{cases}
      0 & text{for } y < 0 \
      y & text{for } y in [0,1) \
      1 & text{for } x ge 1
      end{cases}$$



      To obtain the pdf you just differentiate the cdf of Y.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.



        $P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$



        Now you plug in the value $sqrt x$ into the cdf of $Y$:



        $$F_Y(y)=begin{cases}
        0 & text{for } y < 0 \
        y & text{for } y in [0,1) \
        1 & text{for } x ge 1
        end{cases}$$



        To obtain the pdf you just differentiate the cdf of Y.






        share|cite|improve this answer









        $endgroup$



        You have $X=Y^2$, where $Y sim U(0,1)$. Now we want the cdf of $X$.



        $P(X<x)=P(Y^2<x)=P(Y<sqrt x)=F_Y(sqrt x)$



        Now you plug in the value $sqrt x$ into the cdf of $Y$:



        $$F_Y(y)=begin{cases}
        0 & text{for } y < 0 \
        y & text{for } y in [0,1) \
        1 & text{for } x ge 1
        end{cases}$$



        To obtain the pdf you just differentiate the cdf of Y.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 17:48









        callculuscallculus

        18.2k31427




        18.2k31427






























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