# Conditional expectation of a vector

Suppose we have two random vectors $$X=(X_1,X_2)^T$$ and $$Y=(Y_1,\dots,Y_n)^T$$. I wish to find a simple definition or formula for

$$E_{X|Y=y}[X]$$

Intuitively, I think the following is correct:

$$E_{X|Y=y}[X]=\begin{pmatrix}E_{X_1|Y=y}[X_1]\\E_{X_2|Y=y}[X_2]\end{pmatrix} \tag{1}$$

Is this right? I imagine another possible definition could be

$$E_{X|Y=y}[X]=\begin{pmatrix}E_{X|Y=y}[X_1]\\E_{X|Y=y}[X_2]\end{pmatrix} \tag{2}$$

But I don't think (2) is correct. I say this because, in the case of discrete $$X$$, (3) is more intuitive to me than (4):

$$\big[E_{X|Y=y}[X]\big]_1=\sum_{x_1}x_1p_{X_1|Y}(x_1|y)=\sum_{x_1}x_1P(X_1=x_1|Y=y) \tag{3}$$

$$\big[E_{X|Y=y}[X]\big]_1=\sum_{x}x_1p_{X|Y}(x|y)=\sum_{x}x_1P(X=x|Y=y) \tag{4}$$

• Why are some vectors row vectors and others column vectors? Jan 5, 2019 at 3:39
• Thank you, @DilipSarwate. I have fixed this. They're all column vectors now. Jan 5, 2019 at 3:41

(1) is correct and intuitive; while (2) is also correct because you can calculate $$E[X_1]$$ using the density of $$X_1$$ or the joint density of $$X_1$$ and $$X_2$$, e.g. (in your notation)
\begin{align} E_{X|Y=y}[X_1] &= \sum_x{x_1P(X=x|Y=y)} \\ &= \sum_{x_1}\sum_{x_2}{x_1P(X_1=x_1,X_2=x_2|Y=y)} \\ &= \sum_{x_1}x_1\sum_{x_2}{P(X_1=x_1,X_2=x_2|Y=y)} \\ &= \sum_{x_1}{x_1}P(X_1=x_1|Y=y) \\ &= E_{X_1|Y=y}[X_1] \end{align}