Stuck at defining the borders for a probability density function












0












$begingroup$


So I'm given a continuous random variable $X$ with a probability density function: $p(x)= frac{x+1}{2}$ when $-1 < x< 1$ and the random variable $Y=3X^{2}+1$. I need to find the probability density function for $Y$ however I'm stuck. In order in the statement $y1<y<y2$ to find $y1$ and $y2$ I need to solve the equations: $y1=3*(-1)^2+1$ and $y2 = 3*(1)^{2}+1$, both are being equaled to 4 so I only have one border.



Any ideas?






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    $Pr [Y=y] = Pr [3X^2 +1 =y]= Pr[X= pm frac{y-1}{3}] = Pr [X= frac{y-1}{3}] + Pr[X= - frac{y-1}{3}]$. Where $y in [1,4]$
    $endgroup$
    – mm8511
    Feb 1 at 21:16








  • 1




    $begingroup$
    @mm8511 All your probabilities are $0$.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 23:54










  • $begingroup$
    @KaviRamaMurthy Not really, but i did forget a square root
    $endgroup$
    – mm8511
    Feb 1 at 23:57






  • 1




    $begingroup$
    @mm8511 OP is talking about random variables with densities and you are thinking of them as discrete vaiables.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 23:57
















0












$begingroup$


So I'm given a continuous random variable $X$ with a probability density function: $p(x)= frac{x+1}{2}$ when $-1 < x< 1$ and the random variable $Y=3X^{2}+1$. I need to find the probability density function for $Y$ however I'm stuck. In order in the statement $y1<y<y2$ to find $y1$ and $y2$ I need to solve the equations: $y1=3*(-1)^2+1$ and $y2 = 3*(1)^{2}+1$, both are being equaled to 4 so I only have one border.



Any ideas?






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    $Pr [Y=y] = Pr [3X^2 +1 =y]= Pr[X= pm frac{y-1}{3}] = Pr [X= frac{y-1}{3}] + Pr[X= - frac{y-1}{3}]$. Where $y in [1,4]$
    $endgroup$
    – mm8511
    Feb 1 at 21:16








  • 1




    $begingroup$
    @mm8511 All your probabilities are $0$.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 23:54










  • $begingroup$
    @KaviRamaMurthy Not really, but i did forget a square root
    $endgroup$
    – mm8511
    Feb 1 at 23:57






  • 1




    $begingroup$
    @mm8511 OP is talking about random variables with densities and you are thinking of them as discrete vaiables.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 23:57














0












0








0





$begingroup$


So I'm given a continuous random variable $X$ with a probability density function: $p(x)= frac{x+1}{2}$ when $-1 < x< 1$ and the random variable $Y=3X^{2}+1$. I need to find the probability density function for $Y$ however I'm stuck. In order in the statement $y1<y<y2$ to find $y1$ and $y2$ I need to solve the equations: $y1=3*(-1)^2+1$ and $y2 = 3*(1)^{2}+1$, both are being equaled to 4 so I only have one border.



Any ideas?






probability







share|cite|improve this question









$endgroup$




So I'm given a continuous random variable $X$ with a probability density function: $p(x)= frac{x+1}{2}$ when $-1 < x< 1$ and the random variable $Y=3X^{2}+1$. I need to find the probability density function for $Y$ however I'm stuck. In order in the statement $y1<y<y2$ to find $y1$ and $y2$ I need to solve the equations: $y1=3*(-1)^2+1$ and $y2 = 3*(1)^{2}+1$, both are being equaled to 4 so I only have one border.



Any ideas?






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 1 at 21:06









David MasonDavid Mason

587




587












  • $begingroup$
    $Pr [Y=y] = Pr [3X^2 +1 =y]= Pr[X= pm frac{y-1}{3}] = Pr [X= frac{y-1}{3}] + Pr[X= - frac{y-1}{3}]$. Where $y in [1,4]$
    $endgroup$
    – mm8511
    Feb 1 at 21:16








  • 1




    $begingroup$
    @mm8511 All your probabilities are $0$.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 23:54










  • $begingroup$
    @KaviRamaMurthy Not really, but i did forget a square root
    $endgroup$
    – mm8511
    Feb 1 at 23:57






  • 1




    $begingroup$
    @mm8511 OP is talking about random variables with densities and you are thinking of them as discrete vaiables.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 23:57


















  • $begingroup$
    $Pr [Y=y] = Pr [3X^2 +1 =y]= Pr[X= pm frac{y-1}{3}] = Pr [X= frac{y-1}{3}] + Pr[X= - frac{y-1}{3}]$. Where $y in [1,4]$
    $endgroup$
    – mm8511
    Feb 1 at 21:16








  • 1




    $begingroup$
    @mm8511 All your probabilities are $0$.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 23:54










  • $begingroup$
    @KaviRamaMurthy Not really, but i did forget a square root
    $endgroup$
    – mm8511
    Feb 1 at 23:57






  • 1




    $begingroup$
    @mm8511 OP is talking about random variables with densities and you are thinking of them as discrete vaiables.
    $endgroup$
    – Kavi Rama Murthy
    Feb 1 at 23:57
















$begingroup$
$Pr [Y=y] = Pr [3X^2 +1 =y]= Pr[X= pm frac{y-1}{3}] = Pr [X= frac{y-1}{3}] + Pr[X= - frac{y-1}{3}]$. Where $y in [1,4]$
$endgroup$
– mm8511
Feb 1 at 21:16






$begingroup$
$Pr [Y=y] = Pr [3X^2 +1 =y]= Pr[X= pm frac{y-1}{3}] = Pr [X= frac{y-1}{3}] + Pr[X= - frac{y-1}{3}]$. Where $y in [1,4]$
$endgroup$
– mm8511
Feb 1 at 21:16






1




1




$begingroup$
@mm8511 All your probabilities are $0$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:54




$begingroup$
@mm8511 All your probabilities are $0$.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:54












$begingroup$
@KaviRamaMurthy Not really, but i did forget a square root
$endgroup$
– mm8511
Feb 1 at 23:57




$begingroup$
@KaviRamaMurthy Not really, but i did forget a square root
$endgroup$
– mm8511
Feb 1 at 23:57




1




1




$begingroup$
@mm8511 OP is talking about random variables with densities and you are thinking of them as discrete vaiables.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:57




$begingroup$
@mm8511 OP is talking about random variables with densities and you are thinking of them as discrete vaiables.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:57










2 Answers
2






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$begingroup$

$P{3X^{2}+1leq y}=P{-sqrt {frac {y-1} 3} leq X leq sqrt {frac {y-1} 3}}=int_{-sqrt {frac {y-1} 3}}^{sqrt {frac {y-1} 3}} frac {1+x} 2, dx$ provided $1 <y <4$ and the probability is $1$ for $y >4$. I leave it to you to differentiate and get the density function.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Since $|X|le 1$ a.s., we have $Y=3X^2+1in [ 3cdot 0^2+1, 3cdot 1 ^2+1]=[1,4]$ a.s. . Now let $y$ be a value in $[1,4)$ and let $epsilon>0$ be so that $[y,y+epsilon]$ is included in $[1,4]$. Then:
    $$
    begin{aligned}
    Bbb P(Yin[y,y+epsilon])
    &=
    Bbb P(yle Yle y+epsilon)
    \
    &=
    Bbb P(yle 3X^2+1le y+epsilon)
    \
    &=
    Bbb P(y-1le 3X^2le y-1+epsilon)
    \
    &=
    Bbb Pleft(frac 13(y-1)le X^2le frac 13(y-1+epsilon)right)
    \[2mm]
    &=
    Bbb Pleft(-sqrt{frac 13(y-1)}ge Xge -sqrt{frac 13(y-1+epsilon)}right)
    \
    &+
    Bbb Pleft(+sqrt{frac 13(y-1)}le Xle +sqrt{frac 13(y-1+epsilon)}right)
    \[2mm]
    &=
    int
    _{-sqrt{(y-1+epsilon)/3}}
    ^{-sqrt{(y-1)/3}}
    frac 12(x+1); dx
    \
    &+
    int
    ^{sqrt{(y-1+epsilon)/3}}
    _{sqrt{(y-1)/3}}
    frac 12(x+1); dx
    \[2mm]
    &=
    left[ frac 14(x+1)^2 right]
    _{-sqrt{(y-1+epsilon)/3}}
    ^{-sqrt{(y-1)/3}}
    +
    left[ frac 14(x+1)^2 right]
    ^{sqrt{(y-1+epsilon)/3}}
    _{sqrt{(y-1)/3}}
    \[2mm]
    \&text{and let us denote by $F(x)$ the function $(x+1)^2/4$,}
    \&text{and by $G(x)$ the function $G(x)=F(x)-F(-x)=frac 14(x^2+2x+1)-frac 14(x^2-2x+1)=x$,}
    \[2mm]
    &=
    F(+sqrt{(y-1+epsilon)/3})
    -
    F(-sqrt{(y-1+epsilon)/3})
    \
    &qquad
    -F(+sqrt{(y-1)/3})
    +
    F(-sqrt{(y-1)/3})
    \
    &=
    G(sqrt{(y-1+epsilon)/3}
    -
    G(sqrt{(y-1)/3}
    \
    &=
    sqrt{(y-1+epsilon)/3}
    -
    sqrt{(y-1)/3}
    .end{aligned}
    $$
    And now we can put the hands on the density (which exists, something based on the above computation shows it),
    it is a function
    $rho(y)$ with support in $[1,4]$, and for $y$ in this interval it is given by
    $$
    rho(y)=
    lim_{epsilonsearrow 0}frac 1{epsilon}
    Bbb P(Yin[y,y+epsilon])
    =
    fracpartial{partial epsilon}
    Big(
    sqrt{(y-1+epsilon)/3}
    Big)
    Big|_{epsilon=0}
    =frac 1{2sqrt{3(y-1)}}
    .
    $$
    I always check in such situations...



    sage: var('y');
    
sage: integral( 1/2/sqrt((y-1)*3), y, 1, 4)
    
1





    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      0












      $begingroup$

      $P{3X^{2}+1leq y}=P{-sqrt {frac {y-1} 3} leq X leq sqrt {frac {y-1} 3}}=int_{-sqrt {frac {y-1} 3}}^{sqrt {frac {y-1} 3}} frac {1+x} 2, dx$ provided $1 <y <4$ and the probability is $1$ for $y >4$. I leave it to you to differentiate and get the density function.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $P{3X^{2}+1leq y}=P{-sqrt {frac {y-1} 3} leq X leq sqrt {frac {y-1} 3}}=int_{-sqrt {frac {y-1} 3}}^{sqrt {frac {y-1} 3}} frac {1+x} 2, dx$ provided $1 <y <4$ and the probability is $1$ for $y >4$. I leave it to you to differentiate and get the density function.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $P{3X^{2}+1leq y}=P{-sqrt {frac {y-1} 3} leq X leq sqrt {frac {y-1} 3}}=int_{-sqrt {frac {y-1} 3}}^{sqrt {frac {y-1} 3}} frac {1+x} 2, dx$ provided $1 <y <4$ and the probability is $1$ for $y >4$. I leave it to you to differentiate and get the density function.






          share|cite|improve this answer









          $endgroup$



          $P{3X^{2}+1leq y}=P{-sqrt {frac {y-1} 3} leq X leq sqrt {frac {y-1} 3}}=int_{-sqrt {frac {y-1} 3}}^{sqrt {frac {y-1} 3}} frac {1+x} 2, dx$ provided $1 <y <4$ and the probability is $1$ for $y >4$. I leave it to you to differentiate and get the density function.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 1 at 23:52









          Kavi Rama MurthyKavi Rama Murthy

          74.2k53270




          74.2k53270























              0












              $begingroup$

              Since $|X|le 1$ a.s., we have $Y=3X^2+1in [ 3cdot 0^2+1, 3cdot 1 ^2+1]=[1,4]$ a.s. . Now let $y$ be a value in $[1,4)$ and let $epsilon>0$ be so that $[y,y+epsilon]$ is included in $[1,4]$. Then:
              $$
              begin{aligned}
              Bbb P(Yin[y,y+epsilon])
              &=
              Bbb P(yle Yle y+epsilon)
              \
              &=
              Bbb P(yle 3X^2+1le y+epsilon)
              \
              &=
              Bbb P(y-1le 3X^2le y-1+epsilon)
              \
              &=
              Bbb Pleft(frac 13(y-1)le X^2le frac 13(y-1+epsilon)right)
              \[2mm]
              &=
              Bbb Pleft(-sqrt{frac 13(y-1)}ge Xge -sqrt{frac 13(y-1+epsilon)}right)
              \
              &+
              Bbb Pleft(+sqrt{frac 13(y-1)}le Xle +sqrt{frac 13(y-1+epsilon)}right)
              \[2mm]
              &=
              int
              _{-sqrt{(y-1+epsilon)/3}}
              ^{-sqrt{(y-1)/3}}
              frac 12(x+1); dx
              \
              &+
              int
              ^{sqrt{(y-1+epsilon)/3}}
              _{sqrt{(y-1)/3}}
              frac 12(x+1); dx
              \[2mm]
              &=
              left[ frac 14(x+1)^2 right]
              _{-sqrt{(y-1+epsilon)/3}}
              ^{-sqrt{(y-1)/3}}
              +
              left[ frac 14(x+1)^2 right]
              ^{sqrt{(y-1+epsilon)/3}}
              _{sqrt{(y-1)/3}}
              \[2mm]
              \&text{and let us denote by $F(x)$ the function $(x+1)^2/4$,}
              \&text{and by $G(x)$ the function $G(x)=F(x)-F(-x)=frac 14(x^2+2x+1)-frac 14(x^2-2x+1)=x$,}
              \[2mm]
              &=
              F(+sqrt{(y-1+epsilon)/3})
              -
              F(-sqrt{(y-1+epsilon)/3})
              \
              &qquad
              -F(+sqrt{(y-1)/3})
              +
              F(-sqrt{(y-1)/3})
              \
              &=
              G(sqrt{(y-1+epsilon)/3}
              -
              G(sqrt{(y-1)/3}
              \
              &=
              sqrt{(y-1+epsilon)/3}
              -
              sqrt{(y-1)/3}
              .end{aligned}
              $$
              And now we can put the hands on the density (which exists, something based on the above computation shows it),
              it is a function
              $rho(y)$ with support in $[1,4]$, and for $y$ in this interval it is given by
              $$
              rho(y)=
              lim_{epsilonsearrow 0}frac 1{epsilon}
              Bbb P(Yin[y,y+epsilon])
              =
              fracpartial{partial epsilon}
              Big(
              sqrt{(y-1+epsilon)/3}
              Big)
              Big|_{epsilon=0}
              =frac 1{2sqrt{3(y-1)}}
              .
              $$
              I always check in such situations...



              sage: var('y');
              
sage: integral( 1/2/sqrt((y-1)*3), y, 1, 4)
              
1





              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since $|X|le 1$ a.s., we have $Y=3X^2+1in [ 3cdot 0^2+1, 3cdot 1 ^2+1]=[1,4]$ a.s. . Now let $y$ be a value in $[1,4)$ and let $epsilon>0$ be so that $[y,y+epsilon]$ is included in $[1,4]$. Then:
                $$
                begin{aligned}
                Bbb P(Yin[y,y+epsilon])
                &=
                Bbb P(yle Yle y+epsilon)
                \
                &=
                Bbb P(yle 3X^2+1le y+epsilon)
                \
                &=
                Bbb P(y-1le 3X^2le y-1+epsilon)
                \
                &=
                Bbb Pleft(frac 13(y-1)le X^2le frac 13(y-1+epsilon)right)
                \[2mm]
                &=
                Bbb Pleft(-sqrt{frac 13(y-1)}ge Xge -sqrt{frac 13(y-1+epsilon)}right)
                \
                &+
                Bbb Pleft(+sqrt{frac 13(y-1)}le Xle +sqrt{frac 13(y-1+epsilon)}right)
                \[2mm]
                &=
                int
                _{-sqrt{(y-1+epsilon)/3}}
                ^{-sqrt{(y-1)/3}}
                frac 12(x+1); dx
                \
                &+
                int
                ^{sqrt{(y-1+epsilon)/3}}
                _{sqrt{(y-1)/3}}
                frac 12(x+1); dx
                \[2mm]
                &=
                left[ frac 14(x+1)^2 right]
                _{-sqrt{(y-1+epsilon)/3}}
                ^{-sqrt{(y-1)/3}}
                +
                left[ frac 14(x+1)^2 right]
                ^{sqrt{(y-1+epsilon)/3}}
                _{sqrt{(y-1)/3}}
                \[2mm]
                \&text{and let us denote by $F(x)$ the function $(x+1)^2/4$,}
                \&text{and by $G(x)$ the function $G(x)=F(x)-F(-x)=frac 14(x^2+2x+1)-frac 14(x^2-2x+1)=x$,}
                \[2mm]
                &=
                F(+sqrt{(y-1+epsilon)/3})
                -
                F(-sqrt{(y-1+epsilon)/3})
                \
                &qquad
                -F(+sqrt{(y-1)/3})
                +
                F(-sqrt{(y-1)/3})
                \
                &=
                G(sqrt{(y-1+epsilon)/3}
                -
                G(sqrt{(y-1)/3}
                \
                &=
                sqrt{(y-1+epsilon)/3}
                -
                sqrt{(y-1)/3}
                .end{aligned}
                $$
                And now we can put the hands on the density (which exists, something based on the above computation shows it),
                it is a function
                $rho(y)$ with support in $[1,4]$, and for $y$ in this interval it is given by
                $$
                rho(y)=
                lim_{epsilonsearrow 0}frac 1{epsilon}
                Bbb P(Yin[y,y+epsilon])
                =
                fracpartial{partial epsilon}
                Big(
                sqrt{(y-1+epsilon)/3}
                Big)
                Big|_{epsilon=0}
                =frac 1{2sqrt{3(y-1)}}
                .
                $$
                I always check in such situations...



                sage: var('y');
                
sage: integral( 1/2/sqrt((y-1)*3), y, 1, 4)
                
1





                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since $|X|le 1$ a.s., we have $Y=3X^2+1in [ 3cdot 0^2+1, 3cdot 1 ^2+1]=[1,4]$ a.s. . Now let $y$ be a value in $[1,4)$ and let $epsilon>0$ be so that $[y,y+epsilon]$ is included in $[1,4]$. Then:
                  $$
                  begin{aligned}
                  Bbb P(Yin[y,y+epsilon])
                  &=
                  Bbb P(yle Yle y+epsilon)
                  \
                  &=
                  Bbb P(yle 3X^2+1le y+epsilon)
                  \
                  &=
                  Bbb P(y-1le 3X^2le y-1+epsilon)
                  \
                  &=
                  Bbb Pleft(frac 13(y-1)le X^2le frac 13(y-1+epsilon)right)
                  \[2mm]
                  &=
                  Bbb Pleft(-sqrt{frac 13(y-1)}ge Xge -sqrt{frac 13(y-1+epsilon)}right)
                  \
                  &+
                  Bbb Pleft(+sqrt{frac 13(y-1)}le Xle +sqrt{frac 13(y-1+epsilon)}right)
                  \[2mm]
                  &=
                  int
                  _{-sqrt{(y-1+epsilon)/3}}
                  ^{-sqrt{(y-1)/3}}
                  frac 12(x+1); dx
                  \
                  &+
                  int
                  ^{sqrt{(y-1+epsilon)/3}}
                  _{sqrt{(y-1)/3}}
                  frac 12(x+1); dx
                  \[2mm]
                  &=
                  left[ frac 14(x+1)^2 right]
                  _{-sqrt{(y-1+epsilon)/3}}
                  ^{-sqrt{(y-1)/3}}
                  +
                  left[ frac 14(x+1)^2 right]
                  ^{sqrt{(y-1+epsilon)/3}}
                  _{sqrt{(y-1)/3}}
                  \[2mm]
                  \&text{and let us denote by $F(x)$ the function $(x+1)^2/4$,}
                  \&text{and by $G(x)$ the function $G(x)=F(x)-F(-x)=frac 14(x^2+2x+1)-frac 14(x^2-2x+1)=x$,}
                  \[2mm]
                  &=
                  F(+sqrt{(y-1+epsilon)/3})
                  -
                  F(-sqrt{(y-1+epsilon)/3})
                  \
                  &qquad
                  -F(+sqrt{(y-1)/3})
                  +
                  F(-sqrt{(y-1)/3})
                  \
                  &=
                  G(sqrt{(y-1+epsilon)/3}
                  -
                  G(sqrt{(y-1)/3}
                  \
                  &=
                  sqrt{(y-1+epsilon)/3}
                  -
                  sqrt{(y-1)/3}
                  .end{aligned}
                  $$
                  And now we can put the hands on the density (which exists, something based on the above computation shows it),
                  it is a function
                  $rho(y)$ with support in $[1,4]$, and for $y$ in this interval it is given by
                  $$
                  rho(y)=
                  lim_{epsilonsearrow 0}frac 1{epsilon}
                  Bbb P(Yin[y,y+epsilon])
                  =
                  fracpartial{partial epsilon}
                  Big(
                  sqrt{(y-1+epsilon)/3}
                  Big)
                  Big|_{epsilon=0}
                  =frac 1{2sqrt{3(y-1)}}
                  .
                  $$
                  I always check in such situations...



                  sage: var('y');
                  
sage: integral( 1/2/sqrt((y-1)*3), y, 1, 4)
                  
1





                  share|cite|improve this answer









                  $endgroup$



                  Since $|X|le 1$ a.s., we have $Y=3X^2+1in [ 3cdot 0^2+1, 3cdot 1 ^2+1]=[1,4]$ a.s. . Now let $y$ be a value in $[1,4)$ and let $epsilon>0$ be so that $[y,y+epsilon]$ is included in $[1,4]$. Then:
                  $$
                  begin{aligned}
                  Bbb P(Yin[y,y+epsilon])
                  &=
                  Bbb P(yle Yle y+epsilon)
                  \
                  &=
                  Bbb P(yle 3X^2+1le y+epsilon)
                  \
                  &=
                  Bbb P(y-1le 3X^2le y-1+epsilon)
                  \
                  &=
                  Bbb Pleft(frac 13(y-1)le X^2le frac 13(y-1+epsilon)right)
                  \[2mm]
                  &=
                  Bbb Pleft(-sqrt{frac 13(y-1)}ge Xge -sqrt{frac 13(y-1+epsilon)}right)
                  \
                  &+
                  Bbb Pleft(+sqrt{frac 13(y-1)}le Xle +sqrt{frac 13(y-1+epsilon)}right)
                  \[2mm]
                  &=
                  int
                  _{-sqrt{(y-1+epsilon)/3}}
                  ^{-sqrt{(y-1)/3}}
                  frac 12(x+1); dx
                  \
                  &+
                  int
                  ^{sqrt{(y-1+epsilon)/3}}
                  _{sqrt{(y-1)/3}}
                  frac 12(x+1); dx
                  \[2mm]
                  &=
                  left[ frac 14(x+1)^2 right]
                  _{-sqrt{(y-1+epsilon)/3}}
                  ^{-sqrt{(y-1)/3}}
                  +
                  left[ frac 14(x+1)^2 right]
                  ^{sqrt{(y-1+epsilon)/3}}
                  _{sqrt{(y-1)/3}}
                  \[2mm]
                  \&text{and let us denote by $F(x)$ the function $(x+1)^2/4$,}
                  \&text{and by $G(x)$ the function $G(x)=F(x)-F(-x)=frac 14(x^2+2x+1)-frac 14(x^2-2x+1)=x$,}
                  \[2mm]
                  &=
                  F(+sqrt{(y-1+epsilon)/3})
                  -
                  F(-sqrt{(y-1+epsilon)/3})
                  \
                  &qquad
                  -F(+sqrt{(y-1)/3})
                  +
                  F(-sqrt{(y-1)/3})
                  \
                  &=
                  G(sqrt{(y-1+epsilon)/3}
                  -
                  G(sqrt{(y-1)/3}
                  \
                  &=
                  sqrt{(y-1+epsilon)/3}
                  -
                  sqrt{(y-1)/3}
                  .end{aligned}
                  $$
                  And now we can put the hands on the density (which exists, something based on the above computation shows it),
                  it is a function
                  $rho(y)$ with support in $[1,4]$, and for $y$ in this interval it is given by
                  $$
                  rho(y)=
                  lim_{epsilonsearrow 0}frac 1{epsilon}
                  Bbb P(Yin[y,y+epsilon])
                  =
                  fracpartial{partial epsilon}
                  Big(
                  sqrt{(y-1+epsilon)/3}
                  Big)
                  Big|_{epsilon=0}
                  =frac 1{2sqrt{3(y-1)}}
                  .
                  $$
                  I always check in such situations...



                  sage: var('y');
                  
sage: integral( 1/2/sqrt((y-1)*3), y, 1, 4)
                  
1






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                  answered Feb 4 at 12:12









                  dan_fuleadan_fulea

                  7,1781513




                  7,1781513






























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