$U_1,U_2,…$ i.i.d. $U[0,1]$, $Psim mathrm{Poi}(lambda)$, find $F_{operatorname{min}(U_1,…,U_P)}$












2












$begingroup$



Let $(U_n)_n$ a sequence of random variables i.i.d $U[0,1]$ and let $Psim mathrm{Poi}(lambda)$ a random variable such that $P$ is independent of $(U_n)_n$. Let
$$
\ X=left{begin{matrix}
operatorname{min}{U_1,...,U_P}, Pne 0\
1,P=0
end{matrix}right.
$$
Find $F_X(t)$.




I saw that the solution is
$$
F_X(t)=left{begin{matrix}
0,tleq 0\
1-e^{-lambda t},0leq t<1 \
1, 1leq t
end{matrix}right.\
$$
But my solution is $F_X(t)=e^{-lambda}(e^{lambda t}-1)$ and I don't know where is my mistake:



Obviously for $1leq t$ then $mathbb{P}(Xleq t)=1$ and if $t<0$ then $mathbb{P}(Xleq t)=0$. For $0leq t <1$,
$$
\ mathbb{P}(Xleq t)=mathbb{P}(Xleq t,P=0)+mathbb{P}(Xleq t,P>0)
$$
But if $P=0$ then $X=1$ then $mathbb{P}(Xleq t,P=0)=0$, thus
$$
\ mathbb{P}(Xleq t)=mathbb{P}(Xleq t, P>0)=sum_{k=1}^infty left(prod_{j=1}^kmathbb{P}(U_jleq t)right)mathbb{P}(P=k)=sum_{k=1}^infty {(tlambda)^kover k!}cdot e^{-lambda}
\ =e^{-lambda}sum_{k=0}^infty{(tlambda)^kover k!}-e^{-lambda}= e^{-lambda}cdot e^{tlambda} -e^{-lambda} =e^{-lambda}(e^{lambda t}-1).
$$






probability random-variables uniform-distribution poisson-distribution







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  • 1




    $begingroup$
    You seme to assume that $P=k$ implies $min (U_1,U_2,...,U_P)=U_k$ which is false.
    $endgroup$
    – Kavi Rama Murthy
    Jan 20 at 11:45


















2












$begingroup$



Let $(U_n)_n$ a sequence of random variables i.i.d $U[0,1]$ and let $Psim mathrm{Poi}(lambda)$ a random variable such that $P$ is independent of $(U_n)_n$. Let
$$
\ X=left{begin{matrix}
operatorname{min}{U_1,...,U_P}, Pne 0\
1,P=0
end{matrix}right.
$$
Find $F_X(t)$.




I saw that the solution is
$$
F_X(t)=left{begin{matrix}
0,tleq 0\
1-e^{-lambda t},0leq t<1 \
1, 1leq t
end{matrix}right.\
$$
But my solution is $F_X(t)=e^{-lambda}(e^{lambda t}-1)$ and I don't know where is my mistake:



Obviously for $1leq t$ then $mathbb{P}(Xleq t)=1$ and if $t<0$ then $mathbb{P}(Xleq t)=0$. For $0leq t <1$,
$$
\ mathbb{P}(Xleq t)=mathbb{P}(Xleq t,P=0)+mathbb{P}(Xleq t,P>0)
$$
But if $P=0$ then $X=1$ then $mathbb{P}(Xleq t,P=0)=0$, thus
$$
\ mathbb{P}(Xleq t)=mathbb{P}(Xleq t, P>0)=sum_{k=1}^infty left(prod_{j=1}^kmathbb{P}(U_jleq t)right)mathbb{P}(P=k)=sum_{k=1}^infty {(tlambda)^kover k!}cdot e^{-lambda}
\ =e^{-lambda}sum_{k=0}^infty{(tlambda)^kover k!}-e^{-lambda}= e^{-lambda}cdot e^{tlambda} -e^{-lambda} =e^{-lambda}(e^{lambda t}-1).
$$






probability random-variables uniform-distribution poisson-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You seme to assume that $P=k$ implies $min (U_1,U_2,...,U_P)=U_k$ which is false.
    $endgroup$
    – Kavi Rama Murthy
    Jan 20 at 11:45
















2












2








2





$begingroup$



Let $(U_n)_n$ a sequence of random variables i.i.d $U[0,1]$ and let $Psim mathrm{Poi}(lambda)$ a random variable such that $P$ is independent of $(U_n)_n$. Let
$$
\ X=left{begin{matrix}
operatorname{min}{U_1,...,U_P}, Pne 0\
1,P=0
end{matrix}right.
$$
Find $F_X(t)$.




I saw that the solution is
$$
F_X(t)=left{begin{matrix}
0,tleq 0\
1-e^{-lambda t},0leq t<1 \
1, 1leq t
end{matrix}right.\
$$
But my solution is $F_X(t)=e^{-lambda}(e^{lambda t}-1)$ and I don't know where is my mistake:



Obviously for $1leq t$ then $mathbb{P}(Xleq t)=1$ and if $t<0$ then $mathbb{P}(Xleq t)=0$. For $0leq t <1$,
$$
\ mathbb{P}(Xleq t)=mathbb{P}(Xleq t,P=0)+mathbb{P}(Xleq t,P>0)
$$
But if $P=0$ then $X=1$ then $mathbb{P}(Xleq t,P=0)=0$, thus
$$
\ mathbb{P}(Xleq t)=mathbb{P}(Xleq t, P>0)=sum_{k=1}^infty left(prod_{j=1}^kmathbb{P}(U_jleq t)right)mathbb{P}(P=k)=sum_{k=1}^infty {(tlambda)^kover k!}cdot e^{-lambda}
\ =e^{-lambda}sum_{k=0}^infty{(tlambda)^kover k!}-e^{-lambda}= e^{-lambda}cdot e^{tlambda} -e^{-lambda} =e^{-lambda}(e^{lambda t}-1).
$$






probability random-variables uniform-distribution poisson-distribution







share|cite|improve this question











$endgroup$





Let $(U_n)_n$ a sequence of random variables i.i.d $U[0,1]$ and let $Psim mathrm{Poi}(lambda)$ a random variable such that $P$ is independent of $(U_n)_n$. Let
$$
\ X=left{begin{matrix}
operatorname{min}{U_1,...,U_P}, Pne 0\
1,P=0
end{matrix}right.
$$
Find $F_X(t)$.




I saw that the solution is
$$
F_X(t)=left{begin{matrix}
0,tleq 0\
1-e^{-lambda t},0leq t<1 \
1, 1leq t
end{matrix}right.\
$$
But my solution is $F_X(t)=e^{-lambda}(e^{lambda t}-1)$ and I don't know where is my mistake:



Obviously for $1leq t$ then $mathbb{P}(Xleq t)=1$ and if $t<0$ then $mathbb{P}(Xleq t)=0$. For $0leq t <1$,
$$
\ mathbb{P}(Xleq t)=mathbb{P}(Xleq t,P=0)+mathbb{P}(Xleq t,P>0)
$$
But if $P=0$ then $X=1$ then $mathbb{P}(Xleq t,P=0)=0$, thus
$$
\ mathbb{P}(Xleq t)=mathbb{P}(Xleq t, P>0)=sum_{k=1}^infty left(prod_{j=1}^kmathbb{P}(U_jleq t)right)mathbb{P}(P=k)=sum_{k=1}^infty {(tlambda)^kover k!}cdot e^{-lambda}
\ =e^{-lambda}sum_{k=0}^infty{(tlambda)^kover k!}-e^{-lambda}= e^{-lambda}cdot e^{tlambda} -e^{-lambda} =e^{-lambda}(e^{lambda t}-1).
$$






probability random-variables uniform-distribution poisson-distribution




probability random-variables uniform-distribution poisson-distribution






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edited Feb 2 at 10:27









Davide Giraudo

128k17156268




128k17156268










asked Jan 20 at 10:29









J. DoeJ. Doe

14713




14713








  • 1




    $begingroup$
    You seme to assume that $P=k$ implies $min (U_1,U_2,...,U_P)=U_k$ which is false.
    $endgroup$
    – Kavi Rama Murthy
    Jan 20 at 11:45
















  • 1




    $begingroup$
    You seme to assume that $P=k$ implies $min (U_1,U_2,...,U_P)=U_k$ which is false.
    $endgroup$
    – Kavi Rama Murthy
    Jan 20 at 11:45










1




1




$begingroup$
You seme to assume that $P=k$ implies $min (U_1,U_2,...,U_P)=U_k$ which is false.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 11:45






$begingroup$
You seme to assume that $P=k$ implies $min (U_1,U_2,...,U_P)=U_k$ which is false.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 11:45












1 Answer
1






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oldest

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4












$begingroup$

Since $P(0le Xle 1)=1$, it suffices to show that
$$
Bbb P(X>t)=e^{-lambda t}
$$ for all $tin [0,1)$. Note that
$$begin{eqnarray}
Bbb P(X>t)&=&sum_{k=0}^infty Bbb P(X>t, P=k)\
&=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t, P=k)\
&=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t)Bbb P( P=k)\
&=&sum_{k=0}^infty Bbb P(U_i>t,forall ile k)Bbb P( P=k)\
&=&sum_{k=0}^infty (1-t)^kBbb P( P=k)\
&=&sum_{k=0}^infty (1-t)^kfrac{lambda^ke^{-lambda}}{k!}\
&=&sum_{k=0}^infty frac{(lambda(1-t))^k e^{-lambda}}{k!}=e^{lambda(1-t)}e^{-lambda}=e^{-lambda t}
end{eqnarray}$$ as desired.






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    4












    $begingroup$

    Since $P(0le Xle 1)=1$, it suffices to show that
    $$
    Bbb P(X>t)=e^{-lambda t}
    $$     for all $tin [0,1)$. Note that
    $$begin{eqnarray}
    Bbb P(X>t)&=&sum_{k=0}^infty Bbb P(X>t, P=k)\
    &=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t, P=k)\
    &=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t)Bbb P( P=k)\
    &=&sum_{k=0}^infty Bbb P(U_i>t,forall ile k)Bbb P( P=k)\
    &=&sum_{k=0}^infty (1-t)^kBbb P( P=k)\
    &=&sum_{k=0}^infty (1-t)^kfrac{lambda^ke^{-lambda}}{k!}\
    &=&sum_{k=0}^infty frac{(lambda(1-t))^k e^{-lambda}}{k!}=e^{lambda(1-t)}e^{-lambda}=e^{-lambda t}
    end{eqnarray}$$     as desired.






    share|cite|improve this answer









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      4












      $begingroup$

      Since $P(0le Xle 1)=1$, it suffices to show that
      $$
      Bbb P(X>t)=e^{-lambda t}
      $$       for all $tin [0,1)$. Note that
      $$begin{eqnarray}
      Bbb P(X>t)&=&sum_{k=0}^infty Bbb P(X>t, P=k)\
      &=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t, P=k)\
      &=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t)Bbb P( P=k)\
      &=&sum_{k=0}^infty Bbb P(U_i>t,forall ile k)Bbb P( P=k)\
      &=&sum_{k=0}^infty (1-t)^kBbb P( P=k)\
      &=&sum_{k=0}^infty (1-t)^kfrac{lambda^ke^{-lambda}}{k!}\
      &=&sum_{k=0}^infty frac{(lambda(1-t))^k e^{-lambda}}{k!}=e^{lambda(1-t)}e^{-lambda}=e^{-lambda t}
      end{eqnarray}$$       as desired.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Since $P(0le Xle 1)=1$, it suffices to show that
        $$
        Bbb P(X>t)=e^{-lambda t}
        $$         for all $tin [0,1)$. Note that
        $$begin{eqnarray}
        Bbb P(X>t)&=&sum_{k=0}^infty Bbb P(X>t, P=k)\
        &=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t, P=k)\
        &=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t)Bbb P( P=k)\
        &=&sum_{k=0}^infty Bbb P(U_i>t,forall ile k)Bbb P( P=k)\
        &=&sum_{k=0}^infty (1-t)^kBbb P( P=k)\
        &=&sum_{k=0}^infty (1-t)^kfrac{lambda^ke^{-lambda}}{k!}\
        &=&sum_{k=0}^infty frac{(lambda(1-t))^k e^{-lambda}}{k!}=e^{lambda(1-t)}e^{-lambda}=e^{-lambda t}
        end{eqnarray}$$         as desired.






        share|cite|improve this answer









        $endgroup$



        Since $P(0le Xle 1)=1$, it suffices to show that
        $$
        Bbb P(X>t)=e^{-lambda t}
        $$         for all $tin [0,1)$. Note that
        $$begin{eqnarray}
        Bbb P(X>t)&=&sum_{k=0}^infty Bbb P(X>t, P=k)\
        &=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t, P=k)\
        &=&sum_{k=0}^infty Bbb P(min{U_1,ldots,U_k}>t)Bbb P( P=k)\
        &=&sum_{k=0}^infty Bbb P(U_i>t,forall ile k)Bbb P( P=k)\
        &=&sum_{k=0}^infty (1-t)^kBbb P( P=k)\
        &=&sum_{k=0}^infty (1-t)^kfrac{lambda^ke^{-lambda}}{k!}\
        &=&sum_{k=0}^infty frac{(lambda(1-t))^k e^{-lambda}}{k!}=e^{lambda(1-t)}e^{-lambda}=e^{-lambda t}
        end{eqnarray}$$         as desired.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 20 at 11:43









        SongSong

        18.6k21651




        18.6k21651






























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