Show: $forall k in Bbb{Z}:P(X+Y=k)= sum_{lin Bbb{Z}}P(X=l)cdot P(Y=k-l)$Inequality with $|cdot|_p$...
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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?
$begingroup$
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
edited Mar 14 at 21:42
MarianD
1,5891617
asked Mar 14 at 21:19
asddfasddf
820720
$endgroup$
add a comment |
$begingroup$
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
edited Mar 14 at 21:42
MarianD
1,5891617
asked Mar 14 at 21:19
asddfasddf
820720
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Mar 14 at 21:42
MarianD
1,5891617
asked Mar 14 at 21:19
asddfasddf
820720
$endgroup$
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
edited Mar 14 at 21:42
MarianD
1,5891617
asked Mar 14 at 21:19
asddfasddf
820720
edited Mar 14 at 21:42
MarianD
1,5891617
asked Mar 14 at 21:19
asddfasddf
820720
edited Mar 14 at 21:42
MarianD
1,5891617
edited Mar 14 at 21:42
MarianD
1,5891617
edited Mar 14 at 21:42
MarianD
1,5891617
1,5891617
asked Mar 14 at 21:19
asddfasddf
820720
asked Mar 14 at 21:19
asddfasddf
820720
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 14 at 21:31
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,55938
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Mar 14 at 21:31
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,55938
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 14 at 21:31
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,55938
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 14 at 21:31
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,55938
$endgroup$
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.
answered Mar 14 at 21:31
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,55938
answered Mar 14 at 21:31
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,55938
answered Mar 14 at 21:31
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,55938
answered Mar 14 at 21:31
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,55938
2,55938
add a comment |
add a comment |
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$begingroup$
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