Is expected value $E[X_1;X_1leq X_2]= E[X_1;X_1< X_2]$Find $E(X_1|X_2leq x_2, X_3leq x_3)$ where $(X_1,X_2,X_3)$ is multivariate normalConditional expectation: $E[X_1 X_2mid X_1 + X_2 X_3]$Probability and expectation of three ordered random variablesFind the value of $mathbbE(X_1+X_2+ldots+X_N)$ of i.i.d random variables $X_i$s.Verifying calculation inside an expected value problemLet $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(fracX_1+X_2X_1+X_2+X_3$)?Throw a dice-expected value.Finding conditional expectation $E[X_1 | X_2 = x_2]$Expected value- $E(1/3)^t$Expected value of $Z=X_1+X_2$ if $X_1<X_3$.and $Z=X_1$ if $X_3leq X_1$
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Is expected value $E[X_1;X_1leq X_2]= E[X_1;X_1
Find $E(X_1|X_2leq x_2, X_3leq x_3)$ where $(X_1,X_2,X_3)$ is multivariate normalConditional expectation: $E[X_1 X_2mid X_1 + X_2 X_3]$Probability and expectation of three ordered random variablesFind the value of $mathbbE(X_1+X_2+ldots+X_N)$ of i.i.d random variables $X_i$s.Verifying calculation inside an expected value problemLet $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(fracX_1+X_2X_1+X_2+X_3$)?Throw a dice-expected value.Finding conditional expectation $E[X_1 | X_2 = x_2]$Expected value- $E(1/3)^t$Expected value of $Z=X_1+X_2$ if $X_1<X_3$.and $Z=X_1$ if $X_3leq X_1$
$begingroup$
Let $X_1, X_2$ two independent random variables with PDF $f_X_i(x_i)$.
Is this formula is true
$$E[X_1;X_1leq X_2]=
int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
I am asking if the expected value
$$E[X_1;X_1leq X_2]= E[X_1;X_1< X_2]$$,
or
$$E[X_1;X_1< X_2]=E[X_1]-E[X_1;X_1leq X_2].$$
Thanks
conditional-expectation expected-value
asked 2 days ago
MonirMonir
368
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2$ two independent random variables with PDF $f_X_i(x_i)$.
Is this formula is true
$$E[X_1;X_1leq X_2]=
int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
I am asking if the expected value
$$E[X_1;X_1leq X_2]= E[X_1;X_1< X_2]$$,
or
$$E[X_1;X_1< X_2]=E[X_1]-E[X_1;X_1leq X_2].$$
Thanks
conditional-expectation expected-value
asked 2 days ago
MonirMonir
368
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2$ two independent random variables with PDF $f_X_i(x_i)$.
Is this formula is true
$$E[X_1;X_1leq X_2]=
int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
I am asking if the expected value
$$E[X_1;X_1leq X_2]= E[X_1;X_1< X_2]$$,
or
$$E[X_1;X_1< X_2]=E[X_1]-E[X_1;X_1leq X_2].$$
Thanks
conditional-expectation expected-value
asked 2 days ago
MonirMonir
368
$endgroup$
Let $X_1, X_2$ two independent random variables with PDF $f_X_i(x_i)$.
Is this formula is true
$$E[X_1;X_1leq X_2]=
int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
I am asking if the expected value
$$E[X_1;X_1leq X_2]= E[X_1;X_1< X_2]$$,
or
$$E[X_1;X_1< X_2]=E[X_1]-E[X_1;X_1leq X_2].$$
Thanks
conditional-expectation expected-value
conditional-expectation expected-value
asked 2 days ago
MonirMonir
368
asked 2 days ago
MonirMonir
368
asked 2 days ago
MonirMonir
368
asked 2 days ago
MonirMonir
368
asked 2 days ago
MonirMonir
368
368
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
$begingroup$
Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
$endgroup$
– Monir
2 days ago
$begingroup$
@Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Ok, thank you Prof Kavi Rama Murthy.
$endgroup$
– Monir
2 days ago
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
$begingroup$
Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
$endgroup$
– Monir
2 days ago
$begingroup$
@Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Ok, thank you Prof Kavi Rama Murthy.
$endgroup$
– Monir
2 days ago
add a comment |
$begingroup$
If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
$begingroup$
Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
$endgroup$
– Monir
2 days ago
$begingroup$
@Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Ok, thank you Prof Kavi Rama Murthy.
$endgroup$
– Monir
2 days ago
add a comment |
$begingroup$
If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
If $X_1$ and $X_2$ are independent random variables with a densities then $PX_1=X_2=0$ so it is true that $E(X_1;X_1 leq X_2)=E(X_1;X_1 < X_2)$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
66.6k53067
$begingroup$
Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
$endgroup$
– Monir
2 days ago
$begingroup$
@Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Ok, thank you Prof Kavi Rama Murthy.
$endgroup$
– Monir
2 days ago
add a comment |
$begingroup$
Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
$endgroup$
– Monir
2 days ago
$begingroup$
@Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Ok, thank you Prof Kavi Rama Murthy.
$endgroup$
– Monir
2 days ago
$begingroup$
Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
$endgroup$
– Monir
2 days ago
$begingroup$
Ok, is my formula is true $$E[X_1;X_1leq X_2]= int_x_2=0^inftyBig(int_x_1=0^x_2x_1 f_X_1(x_1)d x_1Big)f_X_2(x_2)dx_2$$
$endgroup$
– Monir
2 days ago
$begingroup$
@Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
@Monir It is true provided the random variables are non-negative. In general the integrals start from $-infty$.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Ok, thank you Prof Kavi Rama Murthy.
$endgroup$
– Monir
2 days ago
$begingroup$
Ok, thank you Prof Kavi Rama Murthy.
$endgroup$
– Monir
2 days ago
add a comment |
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