Proof that if $Y=2X+3$ then $f_Y(y)=frac12f_X(frac{y-3}2)$












0












$begingroup$


I want to prove that if probability density function (PDF) of a random variable $X$ is $f_X(x)$ and another random variable $Y$ is defined as $Y=2X+3$, then the PDF of $Y$ is $f_Y(y)=frac{1}{2} f_X(frac{y-3}{2})$.



My Approach:

We Know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$

similarly,$$P(Y leq y)= int_{-infty}^y f_Y(y) , dy$$
Now,$, Y=2X+3$
$$implies P(Y leq y)= P((2X+3) leq y)$$
$$=P(X leq (frac{y-3}{2}))$$
$$implies P(Y leq y)=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy$$
$$implies int_{-infty}^y f_Y(y) , dy=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy $$
Now how can I proceed further to find a relationship between $f_Y$ and $f_X $?






probability-theory probability-distributions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use the change of variable $yto2y+3$ in the very last integral.
    $endgroup$
    – Did
    Jan 8 at 9:56












  • $begingroup$
    if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
    $endgroup$
    – Suresh
    Jan 8 at 10:12










  • $begingroup$
    No we do not get that.
    $endgroup$
    – Did
    Jan 8 at 14:30






  • 1




    $begingroup$
    Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
    $endgroup$
    – Did
    Jan 9 at 6:51






  • 1




    $begingroup$
    You are doing the change of variable, backwards.
    $endgroup$
    – Did
    Jan 10 at 10:38
















0












$begingroup$


I want to prove that if probability density function (PDF) of a random variable $X$ is $f_X(x)$ and another random variable $Y$ is defined as $Y=2X+3$, then the PDF of $Y$ is $f_Y(y)=frac{1}{2} f_X(frac{y-3}{2})$.



My Approach:

We Know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$

similarly,$$P(Y leq y)= int_{-infty}^y f_Y(y) , dy$$
Now,$, Y=2X+3$
$$implies P(Y leq y)= P((2X+3) leq y)$$
$$=P(X leq (frac{y-3}{2}))$$
$$implies P(Y leq y)=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy$$
$$implies int_{-infty}^y f_Y(y) , dy=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy $$
Now how can I proceed further to find a relationship between $f_Y$ and $f_X $?






probability-theory probability-distributions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use the change of variable $yto2y+3$ in the very last integral.
    $endgroup$
    – Did
    Jan 8 at 9:56












  • $begingroup$
    if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
    $endgroup$
    – Suresh
    Jan 8 at 10:12










  • $begingroup$
    No we do not get that.
    $endgroup$
    – Did
    Jan 8 at 14:30






  • 1




    $begingroup$
    Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
    $endgroup$
    – Did
    Jan 9 at 6:51






  • 1




    $begingroup$
    You are doing the change of variable, backwards.
    $endgroup$
    – Did
    Jan 10 at 10:38














0












0








0





$begingroup$


I want to prove that if probability density function (PDF) of a random variable $X$ is $f_X(x)$ and another random variable $Y$ is defined as $Y=2X+3$, then the PDF of $Y$ is $f_Y(y)=frac{1}{2} f_X(frac{y-3}{2})$.



My Approach:

We Know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$

similarly,$$P(Y leq y)= int_{-infty}^y f_Y(y) , dy$$
Now,$, Y=2X+3$
$$implies P(Y leq y)= P((2X+3) leq y)$$
$$=P(X leq (frac{y-3}{2}))$$
$$implies P(Y leq y)=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy$$
$$implies int_{-infty}^y f_Y(y) , dy=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy $$
Now how can I proceed further to find a relationship between $f_Y$ and $f_X $?






probability-theory probability-distributions







share|cite|improve this question











$endgroup$




I want to prove that if probability density function (PDF) of a random variable $X$ is $f_X(x)$ and another random variable $Y$ is defined as $Y=2X+3$, then the PDF of $Y$ is $f_Y(y)=frac{1}{2} f_X(frac{y-3}{2})$.



My Approach:

We Know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$

similarly,$$P(Y leq y)= int_{-infty}^y f_Y(y) , dy$$
Now,$, Y=2X+3$
$$implies P(Y leq y)= P((2X+3) leq y)$$
$$=P(X leq (frac{y-3}{2}))$$
$$implies P(Y leq y)=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy$$
$$implies int_{-infty}^y f_Y(y) , dy=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy $$
Now how can I proceed further to find a relationship between $f_Y$ and $f_X $?






probability-theory probability-distributions




probability-theory probability-distributions






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edited Jan 8 at 9:58









Did

249k23228467




249k23228467










asked Jan 8 at 9:49









SureshSuresh

371110




371110








  • 1




    $begingroup$
    Use the change of variable $yto2y+3$ in the very last integral.
    $endgroup$
    – Did
    Jan 8 at 9:56












  • $begingroup$
    if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
    $endgroup$
    – Suresh
    Jan 8 at 10:12










  • $begingroup$
    No we do not get that.
    $endgroup$
    – Did
    Jan 8 at 14:30






  • 1




    $begingroup$
    Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
    $endgroup$
    – Did
    Jan 9 at 6:51






  • 1




    $begingroup$
    You are doing the change of variable, backwards.
    $endgroup$
    – Did
    Jan 10 at 10:38














  • 1




    $begingroup$
    Use the change of variable $yto2y+3$ in the very last integral.
    $endgroup$
    – Did
    Jan 8 at 9:56












  • $begingroup$
    if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
    $endgroup$
    – Suresh
    Jan 8 at 10:12










  • $begingroup$
    No we do not get that.
    $endgroup$
    – Did
    Jan 8 at 14:30






  • 1




    $begingroup$
    Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
    $endgroup$
    – Did
    Jan 9 at 6:51






  • 1




    $begingroup$
    You are doing the change of variable, backwards.
    $endgroup$
    – Did
    Jan 10 at 10:38








1




1




$begingroup$
Use the change of variable $yto2y+3$ in the very last integral.
$endgroup$
– Did
Jan 8 at 9:56






$begingroup$
Use the change of variable $yto2y+3$ in the very last integral.
$endgroup$
– Did
Jan 8 at 9:56














$begingroup$
if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
$endgroup$
– Suresh
Jan 8 at 10:12




$begingroup$
if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
$endgroup$
– Suresh
Jan 8 at 10:12












$begingroup$
No we do not get that.
$endgroup$
– Did
Jan 8 at 14:30




$begingroup$
No we do not get that.
$endgroup$
– Did
Jan 8 at 14:30




1




1




$begingroup$
Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
$endgroup$
– Did
Jan 9 at 6:51




$begingroup$
Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
$endgroup$
– Did
Jan 9 at 6:51




1




1




$begingroup$
You are doing the change of variable, backwards.
$endgroup$
– Did
Jan 10 at 10:38




$begingroup$
You are doing the change of variable, backwards.
$endgroup$
– Did
Jan 10 at 10:38










2 Answers
2






active

oldest

votes


















1












$begingroup$


"We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



(Notice that I introduce $u$ here in order to avoid the confusing notation).



Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
    $endgroup$
    – Suresh
    Jan 8 at 10:22








  • 1




    $begingroup$
    If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
    $endgroup$
    – drhab
    Jan 8 at 10:25





















2












$begingroup$

$$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



Now, let's differentiate with respect to $y$ by chain rule



$$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$


    "We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




    Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




    You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



    (Notice that I introduce $u$ here in order to avoid the confusing notation).



    Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
      $endgroup$
      – Suresh
      Jan 8 at 10:22








    • 1




      $begingroup$
      If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
      $endgroup$
      – drhab
      Jan 8 at 10:25


















    1












    $begingroup$


    "We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




    Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




    You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



    (Notice that I introduce $u$ here in order to avoid the confusing notation).



    Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
      $endgroup$
      – Suresh
      Jan 8 at 10:22








    • 1




      $begingroup$
      If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
      $endgroup$
      – drhab
      Jan 8 at 10:25
















    1












    1








    1





    $begingroup$


    "We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




    Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




    You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



    (Notice that I introduce $u$ here in order to avoid the confusing notation).



    Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$






    share|cite|improve this answer









    $endgroup$




    "We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




    Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




    You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



    (Notice that I introduce $u$ here in order to avoid the confusing notation).



    Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 10:14









    drhabdrhab

    104k545136




    104k545136












    • $begingroup$
      Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
      $endgroup$
      – Suresh
      Jan 8 at 10:22








    • 1




      $begingroup$
      If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
      $endgroup$
      – drhab
      Jan 8 at 10:25




















    • $begingroup$
      Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
      $endgroup$
      – Suresh
      Jan 8 at 10:22








    • 1




      $begingroup$
      If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
      $endgroup$
      – drhab
      Jan 8 at 10:25


















    $begingroup$
    Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
    $endgroup$
    – Suresh
    Jan 8 at 10:22






    $begingroup$
    Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
    $endgroup$
    – Suresh
    Jan 8 at 10:22






    1




    1




    $begingroup$
    If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
    $endgroup$
    – drhab
    Jan 8 at 10:25






    $begingroup$
    If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
    $endgroup$
    – drhab
    Jan 8 at 10:25













    2












    $begingroup$

    $$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



    Now, let's differentiate with respect to $y$ by chain rule



    $$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



    that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      $$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



      Now, let's differentiate with respect to $y$ by chain rule



      $$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



      that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        $$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



        Now, let's differentiate with respect to $y$ by chain rule



        $$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



        that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$






        share|cite|improve this answer









        $endgroup$



        $$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



        Now, let's differentiate with respect to $y$ by chain rule



        $$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



        that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 9:57









        Siong Thye GohSiong Thye Goh

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