Random Variable Transformation normal/binomial












0












$begingroup$


I have the following problem I can not solve:



We have two indipendent random Variables given by:
$$
X sim N_{(0,1)}
$$
and
$$
Y_p sim B_{(1,p)}
$$
Now I want to show, that $Z_p sim N_{(0,1)}$ $forall p in (0,1)$ , with
$$
Z_p = (-1)^{Y_p}cdot X
$$
Additionally there is a Hint whoch says, that:
$$
mathbb{P}(A) = mathbb{P}(A cap { Y_p = 1}) + mathbb{P}(A cap { Y_p = 0})
$$



I know that a transformation is given by:
$$
F_Z(z) = mathbb{P}(Z_p < z) = mathbb{P}((-1)^{Y_p}cdot X < z)
$$
And this I somehow get my boundarys for the integration. But I cannot figure out how this is done.






random-variables







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$endgroup$












  • $begingroup$
    You should add to your question that $X$ and $Y_p$ are independent.
    $endgroup$
    – drhab
    Dec 16 '18 at 16:10
















0












$begingroup$


I have the following problem I can not solve:



We have two indipendent random Variables given by:
$$
X sim N_{(0,1)}
$$
and
$$
Y_p sim B_{(1,p)}
$$
Now I want to show, that $Z_p sim N_{(0,1)}$ $forall p in (0,1)$ , with
$$
Z_p = (-1)^{Y_p}cdot X
$$
Additionally there is a Hint whoch says, that:
$$
mathbb{P}(A) = mathbb{P}(A cap { Y_p = 1}) + mathbb{P}(A cap { Y_p = 0})
$$



I know that a transformation is given by:
$$
F_Z(z) = mathbb{P}(Z_p < z) = mathbb{P}((-1)^{Y_p}cdot X < z)
$$
And this I somehow get my boundarys for the integration. But I cannot figure out how this is done.






random-variables







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should add to your question that $X$ and $Y_p$ are independent.
    $endgroup$
    – drhab
    Dec 16 '18 at 16:10














0












0








0





$begingroup$


I have the following problem I can not solve:



We have two indipendent random Variables given by:
$$
X sim N_{(0,1)}
$$
and
$$
Y_p sim B_{(1,p)}
$$
Now I want to show, that $Z_p sim N_{(0,1)}$ $forall p in (0,1)$ , with
$$
Z_p = (-1)^{Y_p}cdot X
$$
Additionally there is a Hint whoch says, that:
$$
mathbb{P}(A) = mathbb{P}(A cap { Y_p = 1}) + mathbb{P}(A cap { Y_p = 0})
$$



I know that a transformation is given by:
$$
F_Z(z) = mathbb{P}(Z_p < z) = mathbb{P}((-1)^{Y_p}cdot X < z)
$$
And this I somehow get my boundarys for the integration. But I cannot figure out how this is done.






random-variables







share|cite|improve this question











$endgroup$




I have the following problem I can not solve:



We have two indipendent random Variables given by:
$$
X sim N_{(0,1)}
$$
and
$$
Y_p sim B_{(1,p)}
$$
Now I want to show, that $Z_p sim N_{(0,1)}$ $forall p in (0,1)$ , with
$$
Z_p = (-1)^{Y_p}cdot X
$$
Additionally there is a Hint whoch says, that:
$$
mathbb{P}(A) = mathbb{P}(A cap { Y_p = 1}) + mathbb{P}(A cap { Y_p = 0})
$$



I know that a transformation is given by:
$$
F_Z(z) = mathbb{P}(Z_p < z) = mathbb{P}((-1)^{Y_p}cdot X < z)
$$
And this I somehow get my boundarys for the integration. But I cannot figure out how this is done.






random-variables




random-variables






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share|cite|improve this question













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share|cite|improve this question








edited Dec 16 '18 at 17:44







RedCrayon

















asked Dec 16 '18 at 15:47









RedCrayonRedCrayon

62




62












  • $begingroup$
    You should add to your question that $X$ and $Y_p$ are independent.
    $endgroup$
    – drhab
    Dec 16 '18 at 16:10


















  • $begingroup$
    You should add to your question that $X$ and $Y_p$ are independent.
    $endgroup$
    – drhab
    Dec 16 '18 at 16:10
















$begingroup$
You should add to your question that $X$ and $Y_p$ are independent.
$endgroup$
– drhab
Dec 16 '18 at 16:10




$begingroup$
You should add to your question that $X$ and $Y_p$ are independent.
$endgroup$
– drhab
Dec 16 '18 at 16:10










2 Answers
2






active

oldest

votes


















1












$begingroup$

I preassume that $X$ and $Y_p$ are independent.



For every Borel measurable $A$:
$$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
& =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
& =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
& =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
& =Pleft(Xin Aright)
end{aligned}
$$



The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Welcome to MSE! :-)



    First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



    Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
    begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
    The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
    begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
    The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
    $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
    Hence, $Z_p sim mathrm{N}(0,1)$.






    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












      $begingroup$

      I preassume that $X$ and $Y_p$ are independent.



      For every Borel measurable $A$:
      $$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
      & =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
      & =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
      & =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
      & =Pleft(Xin Aright)
      end{aligned}
      $$



      The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        I preassume that $X$ and $Y_p$ are independent.



        For every Borel measurable $A$:
        $$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
        & =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
        & =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
        & =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
        & =Pleft(Xin Aright)
        end{aligned}
        $$



        The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          I preassume that $X$ and $Y_p$ are independent.



          For every Borel measurable $A$:
          $$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
          & =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
          & =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
          & =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
          & =Pleft(Xin Aright)
          end{aligned}
          $$



          The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.






          share|cite|improve this answer









          $endgroup$



          I preassume that $X$ and $Y_p$ are independent.



          For every Borel measurable $A$:
          $$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
          & =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
          & =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
          & =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
          & =Pleft(Xin Aright)
          end{aligned}
          $$



          The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 16:08









          drhabdrhab

          103k545136




          103k545136























              1












              $begingroup$

              Welcome to MSE! :-)



              First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



              Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
              begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
              The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
              begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
              The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
              $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
              Hence, $Z_p sim mathrm{N}(0,1)$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Welcome to MSE! :-)



                First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



                Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
                begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
                The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
                begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
                The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
                $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
                Hence, $Z_p sim mathrm{N}(0,1)$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Welcome to MSE! :-)



                  First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



                  Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
                  begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
                  The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
                  begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
                  The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
                  $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
                  Hence, $Z_p sim mathrm{N}(0,1)$.






                  share|cite|improve this answer









                  $endgroup$



                  Welcome to MSE! :-)



                  First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



                  Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
                  begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
                  The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
                  begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
                  The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
                  $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
                  Hence, $Z_p sim mathrm{N}(0,1)$.







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Dec 16 '18 at 16:08









                  JonasJonas

                  398212




                  398212






























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