Discrete random variable-expected value The 2019 Stack Overflow Developer Survey Results Are InCDF of a discrete random variable?Show that $E(Y|X)(omega):=sum_xin X(Omega)E(Y|X=x)chi_leftX=xright(omega)$ is a discrete random variableDefinition of Discrete Random VariableThe transformation of a random discrete variable.Expectation of a discrete function of a continuous random variableEquation for the expected value of a discrete random variableDiscrete Probability: Expected value of random variableExpected Value using the indicator random variableContinuous Random Variable: Distribution Function & Expected ValueThrow a dice-expected value.
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Discrete random variable-expected value
The 2019 Stack Overflow Developer Survey Results Are InCDF of a discrete random variable?Show that $E(Y|X)(omega):=sum_xin X(Omega)E(Y|X=x)chi_leftX=xright(omega)$ is a discrete random variableDefinition of Discrete Random VariableThe transformation of a random discrete variable.Expectation of a discrete function of a continuous random variableEquation for the expected value of a discrete random variableDiscrete Probability: Expected value of random variableExpected Value using the indicator random variableContinuous Random Variable: Distribution Function & Expected ValueThrow a dice-expected value.
$begingroup$
For each discrete random variable $X$ and a measurable set $B$, for which $P [B]> 0$,
show that
$E [X | B] = fracE [1_BX]
P [B]$. I have $E [X | B] = fracsum x_i * P(X=x_i cap B)
P [B]$ and I don't know what next.
random-variables conditional-expectation expected-value
edited Mar 30 at 20:20
Ernie060
2,940719
asked Mar 30 at 17:04
KingisKingis
65
$endgroup$
add a comment |
$begingroup$
For each discrete random variable $X$ and a measurable set $B$, for which $P [B]> 0$,
show that
$E [X | B] = fracE [1_BX]
P [B]$. I have $E [X | B] = fracsum x_i * P(X=x_i cap B)
P [B]$ and I don't know what next.
random-variables conditional-expectation expected-value
edited Mar 30 at 20:20
Ernie060
2,940719
asked Mar 30 at 17:04
KingisKingis
65
$endgroup$
add a comment |
$begingroup$
For each discrete random variable $X$ and a measurable set $B$, for which $P [B]> 0$,
show that
$E [X | B] = fracE [1_BX]
P [B]$. I have $E [X | B] = fracsum x_i * P(X=x_i cap B)
P [B]$ and I don't know what next.
random-variables conditional-expectation expected-value
edited Mar 30 at 20:20
Ernie060
2,940719
asked Mar 30 at 17:04
KingisKingis
65
$endgroup$
For each discrete random variable $X$ and a measurable set $B$, for which $P [B]> 0$,
show that
$E [X | B] = fracE [1_BX]
P [B]$. I have $E [X | B] = fracsum x_i * P(X=x_i cap B)
P [B]$ and I don't know what next.
random-variables conditional-expectation expected-value
random-variables conditional-expectation expected-value
edited Mar 30 at 20:20
Ernie060
2,940719
asked Mar 30 at 17:04
KingisKingis
65
edited Mar 30 at 20:20
Ernie060
2,940719
asked Mar 30 at 17:04
KingisKingis
65
edited Mar 30 at 20:20
Ernie060
2,940719
edited Mar 30 at 20:20
Ernie060
2,940719
edited Mar 30 at 20:20
Ernie060
2,940719
2,940719
asked Mar 30 at 17:04
KingisKingis
65
asked Mar 30 at 17:04
KingisKingis
65
asked Mar 30 at 17:04
KingisKingis
65
65
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$fracE[X 1_B]p[B]=fracsum_omega in Omega x times 1_Btimes P(X=x )
P [B]=fracsum_omega in Bcup B^c x times 1_Btimes P(X=x )
P [B]=fracsum_omega in B x times 1_Btimes P(X=x )+0
P [B]=fracsum_omega in B x times P(X=x )
P [B]=fracsum_omega in B x times P(X=x cap omega in B )
P [B]=fracsum_omega in Omega x times P(X=x cap omega in B )
P [B]=sum_omega in Omega x times fracP(X=x cap omega in BP [B] )=sum_omega in Omega x times P(X=x|B )=E(X|B)$
another way
$fracE[X 1_B]p[B]=fracEE[X 1_Bp[B]$
$=fracE(XE[1_Bp[B]=fracE(Xg(X))p[B]$
$=fracsum_x x g(x) p(X=x)p[B]=fracX=x) p(X=x)p[B]$
$=fracsum_x x p(Bp[B]$
$=fracsum_x x p(X=xp[B]$
$=fracB) p[B]$
$=sum_x x p(X=x|B)=E(X|B)$
note that $E[1_B|X]$ is a random variable that it is a function of $X$.
answered Mar 30 at 17:26
masoudmasoud
3228
$endgroup$
$begingroup$
What did you prove?
$endgroup$
– Kingis
Mar 30 at 17:36
$begingroup$
I edited it. is it what you want or not?
$endgroup$
– masoud
Mar 30 at 19:15
$begingroup$
Why do you use $omega$?
$endgroup$
– Kingis
Mar 30 at 19:28
$begingroup$
since $E(X 1_B)=E(X 1_B(omega))$ , $B$ is a measurable set so $B in F$ in the probability space $(Omega, F ,P)$ (so $Bsubset Omega$ such that $Bin F$
$endgroup$
– masoud
Mar 30 at 19:32
$begingroup$
okey but i must write probably $omega$ im my equation when i have X
$endgroup$
– Kingis
Mar 30 at 19:47
|
show 3 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$fracE[X 1_B]p[B]=fracsum_omega in Omega x times 1_Btimes P(X=x )
P [B]=fracsum_omega in Bcup B^c x times 1_Btimes P(X=x )
P [B]=fracsum_omega in B x times 1_Btimes P(X=x )+0
P [B]=fracsum_omega in B x times P(X=x )
P [B]=fracsum_omega in B x times P(X=x cap omega in B )
P [B]=fracsum_omega in Omega x times P(X=x cap omega in B )
P [B]=sum_omega in Omega x times fracP(X=x cap omega in BP [B] )=sum_omega in Omega x times P(X=x|B )=E(X|B)$
another way
$fracE[X 1_B]p[B]=fracEE[X 1_Bp[B]$
$=fracE(XE[1_Bp[B]=fracE(Xg(X))p[B]$
$=fracsum_x x g(x) p(X=x)p[B]=fracX=x) p(X=x)p[B]$
$=fracsum_x x p(Bp[B]$
$=fracsum_x x p(X=xp[B]$
$=fracB) p[B]$
$=sum_x x p(X=x|B)=E(X|B)$
note that $E[1_B|X]$ is a random variable that it is a function of $X$.
answered Mar 30 at 17:26
masoudmasoud
3228
$endgroup$
$begingroup$
What did you prove?
$endgroup$
– Kingis
Mar 30 at 17:36
$begingroup$
I edited it. is it what you want or not?
$endgroup$
– masoud
Mar 30 at 19:15
$begingroup$
Why do you use $omega$?
$endgroup$
– Kingis
Mar 30 at 19:28
$begingroup$
since $E(X 1_B)=E(X 1_B(omega))$ , $B$ is a measurable set so $B in F$ in the probability space $(Omega, F ,P)$ (so $Bsubset Omega$ such that $Bin F$
$endgroup$
– masoud
Mar 30 at 19:32
$begingroup$
okey but i must write probably $omega$ im my equation when i have X
$endgroup$
– Kingis
Mar 30 at 19:47
|
show 3 more comments
$begingroup$
$fracE[X 1_B]p[B]=fracsum_omega in Omega x times 1_Btimes P(X=x )
P [B]=fracsum_omega in Bcup B^c x times 1_Btimes P(X=x )
P [B]=fracsum_omega in B x times 1_Btimes P(X=x )+0
P [B]=fracsum_omega in B x times P(X=x )
P [B]=fracsum_omega in B x times P(X=x cap omega in B )
P [B]=fracsum_omega in Omega x times P(X=x cap omega in B )
P [B]=sum_omega in Omega x times fracP(X=x cap omega in BP [B] )=sum_omega in Omega x times P(X=x|B )=E(X|B)$
another way
$fracE[X 1_B]p[B]=fracEE[X 1_Bp[B]$
$=fracE(XE[1_Bp[B]=fracE(Xg(X))p[B]$
$=fracsum_x x g(x) p(X=x)p[B]=fracX=x) p(X=x)p[B]$
$=fracsum_x x p(Bp[B]$
$=fracsum_x x p(X=xp[B]$
$=fracB) p[B]$
$=sum_x x p(X=x|B)=E(X|B)$
note that $E[1_B|X]$ is a random variable that it is a function of $X$.
answered Mar 30 at 17:26
masoudmasoud
3228
$endgroup$
$begingroup$
What did you prove?
$endgroup$
– Kingis
Mar 30 at 17:36
$begingroup$
I edited it. is it what you want or not?
$endgroup$
– masoud
Mar 30 at 19:15
$begingroup$
Why do you use $omega$?
$endgroup$
– Kingis
Mar 30 at 19:28
$begingroup$
since $E(X 1_B)=E(X 1_B(omega))$ , $B$ is a measurable set so $B in F$ in the probability space $(Omega, F ,P)$ (so $Bsubset Omega$ such that $Bin F$
$endgroup$
– masoud
Mar 30 at 19:32
$begingroup$
okey but i must write probably $omega$ im my equation when i have X
$endgroup$
– Kingis
Mar 30 at 19:47
|
show 3 more comments
$begingroup$
$fracE[X 1_B]p[B]=fracsum_omega in Omega x times 1_Btimes P(X=x )
P [B]=fracsum_omega in Bcup B^c x times 1_Btimes P(X=x )
P [B]=fracsum_omega in B x times 1_Btimes P(X=x )+0
P [B]=fracsum_omega in B x times P(X=x )
P [B]=fracsum_omega in B x times P(X=x cap omega in B )
P [B]=fracsum_omega in Omega x times P(X=x cap omega in B )
P [B]=sum_omega in Omega x times fracP(X=x cap omega in BP [B] )=sum_omega in Omega x times P(X=x|B )=E(X|B)$
another way
$fracE[X 1_B]p[B]=fracEE[X 1_Bp[B]$
$=fracE(XE[1_Bp[B]=fracE(Xg(X))p[B]$
$=fracsum_x x g(x) p(X=x)p[B]=fracX=x) p(X=x)p[B]$
$=fracsum_x x p(Bp[B]$
$=fracsum_x x p(X=xp[B]$
$=fracB) p[B]$
$=sum_x x p(X=x|B)=E(X|B)$
note that $E[1_B|X]$ is a random variable that it is a function of $X$.
answered Mar 30 at 17:26
masoudmasoud
3228
$endgroup$
$fracE[X 1_B]p[B]=fracsum_omega in Omega x times 1_Btimes P(X=x )
P [B]=fracsum_omega in Bcup B^c x times 1_Btimes P(X=x )
P [B]=fracsum_omega in B x times 1_Btimes P(X=x )+0
P [B]=fracsum_omega in B x times P(X=x )
P [B]=fracsum_omega in B x times P(X=x cap omega in B )
P [B]=fracsum_omega in Omega x times P(X=x cap omega in B )
P [B]=sum_omega in Omega x times fracP(X=x cap omega in BP [B] )=sum_omega in Omega x times P(X=x|B )=E(X|B)$
another way
$fracE[X 1_B]p[B]=fracEE[X 1_Bp[B]$
$=fracE(XE[1_Bp[B]=fracE(Xg(X))p[B]$
$=fracsum_x x g(x) p(X=x)p[B]=fracX=x) p(X=x)p[B]$
$=fracsum_x x p(Bp[B]$
$=fracsum_x x p(X=xp[B]$
$=fracB) p[B]$
$=sum_x x p(X=x|B)=E(X|B)$
note that $E[1_B|X]$ is a random variable that it is a function of $X$.
answered Mar 30 at 17:26
masoudmasoud
3228
edited Mar 31 at 13:12
answered Mar 30 at 17:26
masoudmasoud
3228
answered Mar 30 at 17:26
masoudmasoud
3228
answered Mar 30 at 17:26
masoudmasoud
3228
3228
$begingroup$
What did you prove?
$endgroup$
– Kingis
Mar 30 at 17:36
$begingroup$
I edited it. is it what you want or not?
$endgroup$
– masoud
Mar 30 at 19:15
$begingroup$
Why do you use $omega$?
$endgroup$
– Kingis
Mar 30 at 19:28
$begingroup$
since $E(X 1_B)=E(X 1_B(omega))$ , $B$ is a measurable set so $B in F$ in the probability space $(Omega, F ,P)$ (so $Bsubset Omega$ such that $Bin F$
$endgroup$
– masoud
Mar 30 at 19:32
$begingroup$
okey but i must write probably $omega$ im my equation when i have X
$endgroup$
– Kingis
Mar 30 at 19:47
|
show 3 more comments
$begingroup$
What did you prove?
$endgroup$
– Kingis
Mar 30 at 17:36
$begingroup$
I edited it. is it what you want or not?
$endgroup$
– masoud
Mar 30 at 19:15
$begingroup$
Why do you use $omega$?
$endgroup$
– Kingis
Mar 30 at 19:28
$begingroup$
since $E(X 1_B)=E(X 1_B(omega))$ , $B$ is a measurable set so $B in F$ in the probability space $(Omega, F ,P)$ (so $Bsubset Omega$ such that $Bin F$
$endgroup$
– masoud
Mar 30 at 19:32
$begingroup$
okey but i must write probably $omega$ im my equation when i have X
$endgroup$
– Kingis
Mar 30 at 19:47
$begingroup$
What did you prove?
$endgroup$
– Kingis
Mar 30 at 17:36
$begingroup$
What did you prove?
$endgroup$
– Kingis
Mar 30 at 17:36
$begingroup$
I edited it. is it what you want or not?
$endgroup$
– masoud
Mar 30 at 19:15
$begingroup$
I edited it. is it what you want or not?
$endgroup$
– masoud
Mar 30 at 19:15
$begingroup$
Why do you use $omega$?
$endgroup$
– Kingis
Mar 30 at 19:28
$begingroup$
Why do you use $omega$?
$endgroup$
– Kingis
Mar 30 at 19:28
$begingroup$
since $E(X 1_B)=E(X 1_B(omega))$ , $B$ is a measurable set so $B in F$ in the probability space $(Omega, F ,P)$ (so $Bsubset Omega$ such that $Bin F$
$endgroup$
– masoud
Mar 30 at 19:32
$begingroup$
since $E(X 1_B)=E(X 1_B(omega))$ , $B$ is a measurable set so $B in F$ in the probability space $(Omega, F ,P)$ (so $Bsubset Omega$ such that $Bin F$
$endgroup$
– masoud
Mar 30 at 19:32
$begingroup$
okey but i must write probably $omega$ im my equation when i have X
$endgroup$
– Kingis
Mar 30 at 19:47
$begingroup$
okey but i must write probably $omega$ im my equation when i have X
$endgroup$
– Kingis
Mar 30 at 19:47
|
show 3 more comments
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