On Finding Means of Distributions If I have a distribution which depends only on one variable, I usually find the mean by:
(Continuous Case)
$$\mu=\displaystyle \int xf_X(x).dx$$
What happens in the following cases:


*

*Conditional Mean?
$$\mu=\displaystyle \int xf_X(x|y).dx ?$$ 

*Joint Distribution?
$$??f_{X,Y}(x,y)??$$
I have taken a course on this stuff a long time ago but I needed to do some Bayesian Analysis for my research and the Posterior Mean keeps ruining my life.
For instance,
$$f_{\Theta|X}(\theta|x)\propto f_{X|\Theta}(x|\theta)\cdot f_\Theta(\theta)$$
I have $\theta$ as 2 parameters and I have to find posterior mean (which is the expected value of $f_{\Theta|X}$)
 A: Don't forget that the conditional mean should be written $$\mu(y)=\int x f_{X|Y}(x|y)d x$$  and is a function of $y$.

For the posterior mean,
\begin{eqnarray*}
  E \left[ \theta |x \right] & = & \int \theta f_{\Theta |X} \left( \theta |x
  \right) \mathrm{d} \theta\\
  & = & \frac{1}{f_X \left( x \right)} \int \theta f_{\Theta} \left( \theta
  \right) f_{X| \Theta} \left( x \left| \theta \right. \right) \mathrm{d}
  \theta
\end{eqnarray*}
If you have two components, then
\begin{eqnarray*}
  E \left[ \theta |x \right] & = & \left(\begin{array}{c}
    E \left[ \theta_1 |x \right]\\
    E \left[ \theta_2 |x \right]
  \end{array}\right)\\
  & = & \left(\begin{array}{c}
    \int \theta_1 f_{\Theta |X} \left( \theta |x \right) \mathrm{d} \theta_1\\
    \int \theta_2 f_{\Theta |X} \left( \theta |x \right) \mathrm{d} \theta_2
  \end{array}\right)
\end{eqnarray*}
There is nothing special about vectors here. It was implicit, in $\begin{array}{lll}
  E \left[ \theta |x \right] & = & \int \theta f_{\Theta |X} \left( \theta |x
  \right) \mathrm{d} \theta
\end{array}$ that, if $\theta$ is a vector then both sides of the equation are a vector.

For the mean from the joint distribution (and non-rigorously)
\begin{eqnarray*}
  E \left[ X \right] & = & \int xf_{X, Y} \left( x, y \right) \mathrm{d} x
  \mathrm{d} y\\
  & = & \int xf_X \left( x \right) \underbrace{\left( \int f_{Y \left| X
  \right.} \left( y|x \right) \mathrm{d} y \right)}_{= 1} \mathrm{d} x\\
  & = & \int xf_X \left( x \right) \mathrm{d} x
\end{eqnarray*}
So if the object in which you are interested only depends on some marginal variables, you only need to compute the mean with respect to the distribution of those marginal variables.
I hope that things are clearer now!
