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Show: $forall k in Bbb{Z}:P(X+Y=k)= sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l)$


Inequality with $|cdot|_p$ normdistribution of the indicator function of poissonCompute $Bbb E [(n-S_n)^+]$ where $S_n$ is a sum of i.i.d Poisson random variables.Show that the random variables $Y_1$ and $Y_2$ are independentConditional Probability of $k$ independent Poisson DistributionsCounterexample: $Bbb{E}[frac {X_1} {X_1+…+X_n}]nefrac{1}{n}$ if $X_1,…,X_n$ not independentShowing $P(S_m<m, forall 1leq mleq n | S_n)=max{0, 1-S_n/n}$Show that there exist $ b in (0, infty)$ such that $lim_{n to infty} exp(bn) cdot P((1/n)cdot sum_{i=1}^n X_i >a)=0$Expectation of $ | { j in Bbb N : sum_{i=1}^j alpha_i leq t } |$ , $alpha_i$ iid RV on $[0,infty)$Show that $X$, $Y$, and $Z$ are pairwise independent. Are they independent?













0












$begingroup$


Let $X,Y$ be independent random variables with values in $Bbb{Z}$
. Show:
$$forall k in Bbb{Z}:P(X+Y=k)= sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l)$$



My attempt:



begin{aligned}
sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l) &= sum_{lin Bbb{Z}}P(X=l cap Y=k-X) = sum_{lin Bbb{Z}}P(X=l cap Y+X=k)\[1em] &=P(Xin Z cap Y+X=k)=P(Y+X=k)
end{aligned}






probability measure-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    You need to take into account both $X$ and $Y$ have to be positive. In particular the upper limit on the sum has to be $k-1$.
    $endgroup$
    – herb steinberg
    Mar 14 at 21:27












  • $begingroup$
    @herbsteinberg sorry i dont understand. Where exactly is my mistake ?
    $endgroup$
    – asddf
    Mar 14 at 21:34










  • $begingroup$
    I wouldn't say it is a mistake. You just have to realize for $lge k$, $P(Y=k-l)=0$.
    $endgroup$
    – herb steinberg
    Mar 15 at 0:24


















0












$begingroup$


Let $X,Y$ be independent random variables with values in $Bbb{Z}$
. Show:
$$forall k in Bbb{Z}:P(X+Y=k)= sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l)$$



My attempt:



begin{aligned}
sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l) &= sum_{lin Bbb{Z}}P(X=l cap Y=k-X) = sum_{lin Bbb{Z}}P(X=l cap Y+X=k)\[1em] &=P(Xin Z cap Y+X=k)=P(Y+X=k)
end{aligned}






probability measure-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    You need to take into account both $X$ and $Y$ have to be positive. In particular the upper limit on the sum has to be $k-1$.
    $endgroup$
    – herb steinberg
    Mar 14 at 21:27












  • $begingroup$
    @herbsteinberg sorry i dont understand. Where exactly is my mistake ?
    $endgroup$
    – asddf
    Mar 14 at 21:34










  • $begingroup$
    I wouldn't say it is a mistake. You just have to realize for $lge k$, $P(Y=k-l)=0$.
    $endgroup$
    – herb steinberg
    Mar 15 at 0:24
















0












0








0





$begingroup$


Let $X,Y$ be independent random variables with values in $Bbb{Z}$
. Show:
$$forall k in Bbb{Z}:P(X+Y=k)= sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l)$$



My attempt:



begin{aligned}
sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l) &= sum_{lin Bbb{Z}}P(X=l cap Y=k-X) = sum_{lin Bbb{Z}}P(X=l cap Y+X=k)\[1em] &=P(Xin Z cap Y+X=k)=P(Y+X=k)
end{aligned}






probability measure-theory







share|cite|improve this question











$endgroup$




Let $X,Y$ be independent random variables with values in $Bbb{Z}$
. Show:
$$forall k in Bbb{Z}:P(X+Y=k)= sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l)$$



My attempt:



begin{aligned}
sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l) &= sum_{lin Bbb{Z}}P(X=l cap Y=k-X) = sum_{lin Bbb{Z}}P(X=l cap Y+X=k)\[1em] &=P(Xin Z cap Y+X=k)=P(Y+X=k)
end{aligned}






probability measure-theory




probability measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 14 at 21:42









MarianD

1,5891617




1,5891617










asked Mar 14 at 21:19









asddfasddf

820720




820720












  • $begingroup$
    You need to take into account both $X$ and $Y$ have to be positive. In particular the upper limit on the sum has to be $k-1$.
    $endgroup$
    – herb steinberg
    Mar 14 at 21:27












  • $begingroup$
    @herbsteinberg sorry i dont understand. Where exactly is my mistake ?
    $endgroup$
    – asddf
    Mar 14 at 21:34










  • $begingroup$
    I wouldn't say it is a mistake. You just have to realize for $lge k$, $P(Y=k-l)=0$.
    $endgroup$
    – herb steinberg
    Mar 15 at 0:24




















  • $begingroup$
    You need to take into account both $X$ and $Y$ have to be positive. In particular the upper limit on the sum has to be $k-1$.
    $endgroup$
    – herb steinberg
    Mar 14 at 21:27












  • $begingroup$
    @herbsteinberg sorry i dont understand. Where exactly is my mistake ?
    $endgroup$
    – asddf
    Mar 14 at 21:34










  • $begingroup$
    I wouldn't say it is a mistake. You just have to realize for $lge k$, $P(Y=k-l)=0$.
    $endgroup$
    – herb steinberg
    Mar 15 at 0:24


















$begingroup$
You need to take into account both $X$ and $Y$ have to be positive. In particular the upper limit on the sum has to be $k-1$.
$endgroup$
– herb steinberg
Mar 14 at 21:27






$begingroup$
You need to take into account both $X$ and $Y$ have to be positive. In particular the upper limit on the sum has to be $k-1$.
$endgroup$
– herb steinberg
Mar 14 at 21:27














$begingroup$
@herbsteinberg sorry i dont understand. Where exactly is my mistake ?
$endgroup$
– asddf
Mar 14 at 21:34




$begingroup$
@herbsteinberg sorry i dont understand. Where exactly is my mistake ?
$endgroup$
– asddf
Mar 14 at 21:34












$begingroup$
I wouldn't say it is a mistake. You just have to realize for $lge k$, $P(Y=k-l)=0$.
$endgroup$
– herb steinberg
Mar 15 at 0:24






$begingroup$
I wouldn't say it is a mistake. You just have to realize for $lge k$, $P(Y=k-l)=0$.
$endgroup$
– herb steinberg
Mar 15 at 0:24












1 Answer
1






active

oldest

votes


















0












$begingroup$

Let $(Omega,mathcal{F},P)$ be the underlying probability space.
Let $A={omegainOmegamid X(omega)+Y(omega)=k}$. For each $ninmathbb{Z}$,
let $B_{n}={omegainOmegamid X(omega)=n}$ and $C_{n}={omegainOmegamid Y(omega)=k-n}$.
It is routine to check that $A=bigcup_{ninmathbb{Z}}left(B_{n}cap C_{n}right)$.
Moreover, for any $m,ninmathbb{Z}$ with $mneq n$, $(B_{m}cap C_{m})bigcap(B_{n}cap C_{n})=emptyset$.
Therefore
begin{eqnarray*}
P(A) & = & sum_{ninmathbb{Z}}Pleft(B_{n}cap C_{n}right)\
& = & sum_{ninmathbb{Z}}P(B_{n})P(C_{n})\
& = & sum_{ninmathbb{Z}}Pleft(left[X=nright]right)Pleft(left[Y=k-nright]right).
end{eqnarray*}
Note that, the second equality follows from the fact that $X$ and
$Y$ are independent.






share|cite|improve this answer









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    1 Answer
    1






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    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    Let $(Omega,mathcal{F},P)$ be the underlying probability space.
    Let $A={omegainOmegamid X(omega)+Y(omega)=k}$. For each $ninmathbb{Z}$,
    let $B_{n}={omegainOmegamid X(omega)=n}$ and $C_{n}={omegainOmegamid Y(omega)=k-n}$.
    It is routine to check that $A=bigcup_{ninmathbb{Z}}left(B_{n}cap C_{n}right)$.
    Moreover, for any $m,ninmathbb{Z}$ with $mneq n$, $(B_{m}cap C_{m})bigcap(B_{n}cap C_{n})=emptyset$.
    Therefore
    begin{eqnarray*}
    P(A) & = & sum_{ninmathbb{Z}}Pleft(B_{n}cap C_{n}right)\
    & = & sum_{ninmathbb{Z}}P(B_{n})P(C_{n})\
    & = & sum_{ninmathbb{Z}}Pleft(left[X=nright]right)Pleft(left[Y=k-nright]right).
    end{eqnarray*}
    Note that, the second equality follows from the fact that $X$ and
    $Y$ are independent.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $(Omega,mathcal{F},P)$ be the underlying probability space.
      Let $A={omegainOmegamid X(omega)+Y(omega)=k}$. For each $ninmathbb{Z}$,
      let $B_{n}={omegainOmegamid X(omega)=n}$ and $C_{n}={omegainOmegamid Y(omega)=k-n}$.
      It is routine to check that $A=bigcup_{ninmathbb{Z}}left(B_{n}cap C_{n}right)$.
      Moreover, for any $m,ninmathbb{Z}$ with $mneq n$, $(B_{m}cap C_{m})bigcap(B_{n}cap C_{n})=emptyset$.
      Therefore
      begin{eqnarray*}
      P(A) & = & sum_{ninmathbb{Z}}Pleft(B_{n}cap C_{n}right)\
      & = & sum_{ninmathbb{Z}}P(B_{n})P(C_{n})\
      & = & sum_{ninmathbb{Z}}Pleft(left[X=nright]right)Pleft(left[Y=k-nright]right).
      end{eqnarray*}
      Note that, the second equality follows from the fact that $X$ and
      $Y$ are independent.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $(Omega,mathcal{F},P)$ be the underlying probability space.
        Let $A={omegainOmegamid X(omega)+Y(omega)=k}$. For each $ninmathbb{Z}$,
        let $B_{n}={omegainOmegamid X(omega)=n}$ and $C_{n}={omegainOmegamid Y(omega)=k-n}$.
        It is routine to check that $A=bigcup_{ninmathbb{Z}}left(B_{n}cap C_{n}right)$.
        Moreover, for any $m,ninmathbb{Z}$ with $mneq n$, $(B_{m}cap C_{m})bigcap(B_{n}cap C_{n})=emptyset$.
        Therefore
        begin{eqnarray*}
        P(A) & = & sum_{ninmathbb{Z}}Pleft(B_{n}cap C_{n}right)\
        & = & sum_{ninmathbb{Z}}P(B_{n})P(C_{n})\
        & = & sum_{ninmathbb{Z}}Pleft(left[X=nright]right)Pleft(left[Y=k-nright]right).
        end{eqnarray*}
        Note that, the second equality follows from the fact that $X$ and
        $Y$ are independent.






        share|cite|improve this answer









        $endgroup$



        Let $(Omega,mathcal{F},P)$ be the underlying probability space.
        Let $A={omegainOmegamid X(omega)+Y(omega)=k}$. For each $ninmathbb{Z}$,
        let $B_{n}={omegainOmegamid X(omega)=n}$ and $C_{n}={omegainOmegamid Y(omega)=k-n}$.
        It is routine to check that $A=bigcup_{ninmathbb{Z}}left(B_{n}cap C_{n}right)$.
        Moreover, for any $m,ninmathbb{Z}$ with $mneq n$, $(B_{m}cap C_{m})bigcap(B_{n}cap C_{n})=emptyset$.
        Therefore
        begin{eqnarray*}
        P(A) & = & sum_{ninmathbb{Z}}Pleft(B_{n}cap C_{n}right)\
        & = & sum_{ninmathbb{Z}}P(B_{n})P(C_{n})\
        & = & sum_{ninmathbb{Z}}Pleft(left[X=nright]right)Pleft(left[Y=k-nright]right).
        end{eqnarray*}
        Note that, the second equality follows from the fact that $X$ and
        $Y$ are independent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 14 at 21:31









        Danny Pak-Keung ChanDanny Pak-Keung Chan

        2,55938




        2,55938






























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