27 мар. 2019 г. · This is correct: E[XY]=E[E[XY|X]]=E[X[E[Y|X]]=E[X(a+bX)]. By linearity of the expectation operator, we get E[XY]=aE[X]+bE[X2]. |
28 нояб. 2017 г. · ... , Y=μ2+σ2(√1−ρ2V+ρU). This makes (U,V) a standard binormal variable and gives the easy calculation E[XY]=E[μ1μ2+σ1σ2ρU2+aU+bV+cU ... |
24 янв. 2022 г. · You can try with the law of total expectation which gives E[XY]=E[E(XY∣Y)]=E[YE(X∣Y)]. Then E(X∣Y) can be computed ... |
27 мая 2019 г. · By the expression E(XY), we mean the expectation of XY under their joint distribution. I.e., if these both are continuous, we have that. |
10 окт. 2014 г. · The difference between E(X∣Y) and E(X∣Y=y) is that the former is a random variable, whereas the latter is (in some sense) a realization of E(X∣Y). |
17 сент. 2017 г. · E(X|Y) is a random variable, defined for all outcomes of Y, that is equal to the expectation of X given this outcome of Y (E(X|Y=a)). |
26 окт. 2014 г. · You can use independence of X and Y (notice that that is independence with respect to both the variables) to say that E[XY] = E[X]E[Y]. |
27 мар. 2013 г. · Given the expected value of Y and the variance of Y, we can calculate the co-variance of X and Y using the following formula: E[XY] - E[X]*E[Y]. |
16 мар. 2019 г. · I know Var(XY)= E(X^2 Y^2)- (E(XY))^2 and E(XY)= E(X) E(Y) as X, Y are independent but no idea about X^2 and Y^2 are independent or not. – Dev. |
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