Mixing
Truncated conditional expectation of multivariate normal distribution
$begingroup$
Assume that $y_1 = alpha_1 s_1 + u_1$ and $y_2 = alpha_2 s_1 + alpha_3 s_2 + alpha_4 u_1 + alpha_5 u_2$ where $s_1 sim N(0,Sigma_1)$, $s_2 sim N(0,Sigma_2)$, $u_1 sim N(0,sigma^2)$, $u_2 sim N(0,sigma^2)$ with the fact that they are all independently distributed. How can we calculate the following:
$E[s_1 | y_1, y_2, s_1>t]$.
I made the following progress and stuck with the last expression;
$E[s_1 | y_1, y_2, s_1>t] = int f(s_1|y_1, y_2, s_1 >t)s_1 ds_1 = int_{t}^{infty} s_1 f(s_1|y_1, y_2)ds_1$.
Can we get an expression, linear in $y_1$ and $y_2$, from the last expression?
probability probability-distributions conditional-expectation conditional-probability
asked Jan 20 at 5:12
user229519user229519
105
$endgroup$
add a comment |
$begingroup$
Assume that $y_1 = alpha_1 s_1 + u_1$ and $y_2 = alpha_2 s_1 + alpha_3 s_2 + alpha_4 u_1 + alpha_5 u_2$ where $s_1 sim N(0,Sigma_1)$, $s_2 sim N(0,Sigma_2)$, $u_1 sim N(0,sigma^2)$, $u_2 sim N(0,sigma^2)$ with the fact that they are all independently distributed. How can we calculate the following:
$E[s_1 | y_1, y_2, s_1>t]$.
I made the following progress and stuck with the last expression;
$E[s_1 | y_1, y_2, s_1>t] = int f(s_1|y_1, y_2, s_1 >t)s_1 ds_1 = int_{t}^{infty} s_1 f(s_1|y_1, y_2)ds_1$.
Can we get an expression, linear in $y_1$ and $y_2$, from the last expression?
probability probability-distributions conditional-expectation conditional-probability
asked Jan 20 at 5:12
user229519user229519
105
$endgroup$
add a comment |
$begingroup$
Assume that $y_1 = alpha_1 s_1 + u_1$ and $y_2 = alpha_2 s_1 + alpha_3 s_2 + alpha_4 u_1 + alpha_5 u_2$ where $s_1 sim N(0,Sigma_1)$, $s_2 sim N(0,Sigma_2)$, $u_1 sim N(0,sigma^2)$, $u_2 sim N(0,sigma^2)$ with the fact that they are all independently distributed. How can we calculate the following:
$E[s_1 | y_1, y_2, s_1>t]$.
I made the following progress and stuck with the last expression;
$E[s_1 | y_1, y_2, s_1>t] = int f(s_1|y_1, y_2, s_1 >t)s_1 ds_1 = int_{t}^{infty} s_1 f(s_1|y_1, y_2)ds_1$.
Can we get an expression, linear in $y_1$ and $y_2$, from the last expression?
probability probability-distributions conditional-expectation conditional-probability
asked Jan 20 at 5:12
user229519user229519
105
$endgroup$
Assume that $y_1 = alpha_1 s_1 + u_1$ and $y_2 = alpha_2 s_1 + alpha_3 s_2 + alpha_4 u_1 + alpha_5 u_2$ where $s_1 sim N(0,Sigma_1)$, $s_2 sim N(0,Sigma_2)$, $u_1 sim N(0,sigma^2)$, $u_2 sim N(0,sigma^2)$ with the fact that they are all independently distributed. How can we calculate the following:
$E[s_1 | y_1, y_2, s_1>t]$.
I made the following progress and stuck with the last expression;
$E[s_1 | y_1, y_2, s_1>t] = int f(s_1|y_1, y_2, s_1 >t)s_1 ds_1 = int_{t}^{infty} s_1 f(s_1|y_1, y_2)ds_1$.
Can we get an expression, linear in $y_1$ and $y_2$, from the last expression?
probability probability-distributions conditional-expectation conditional-probability
probability probability-distributions conditional-expectation conditional-probability
asked Jan 20 at 5:12
user229519user229519
105
asked Jan 20 at 5:12
user229519user229519
105
asked Jan 20 at 5:12
user229519user229519
105
asked Jan 20 at 5:12
user229519user229519
105
asked Jan 20 at 5:12
user229519user229519
105
105
add a comment |
add a comment |
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