Is following equality true: $$E[X_{1}X_{2}|X_{1}=2]=E[2X_{2}|X_{1}=2]$$
If no how to calculate/simplify this, assuming that we have joint density and marginal densities of those RV and they are not independent.
Also what is generally speaking:
$$E[X|X=x]$$??
On
If $X_1$ and $X_2$ are independent, then the equality $\mathbb E[X_1 \cdot X_2 \mid X_1 = 2] = \mathbb E[2 X_2]$ is true and in this case we can plug $X_1 = 2$ in for $X_1$.
If we do not have this assumption of independence, more generally we have
$$ \mathbb E[X_1 \cdot X_2 \mid X_1 = 2] = \int_{-\infty}^\infty x_1 \cdot 2 \; f_{X_1 X_2 \mid X_2} (x_1 x_2, 2) \; dx_1 = \mathbb E[2 X_2 \mid X_1 = 2] $$
which is a proof
A source for this is: http://math.arizona.edu/~tgk/464_07/cond_exp.pdf
In general, $\mathbb E[X \mid X = x] = x$. Why? Well, consider the quantity $X \mid X = x.$ This is no longer a random variable, it is simply a constant $x$. Qualitatively, an example is, the expected value of a coin flip (0 for tails, 1 for heads), given that the coin flip already happened, is simply the outcome of the coin flip that already happened, which we can observe.
For discrete distributions
Suppose that $X_1$ and $X_2$ take their values from the set $\{1,2,\dots,n\}$ and let the common probability mass function be given: $P(X_1=i\cap X_2=j)$. Also, assume that $P(X_1=2)>0.$
By definition
$$E[X_1X_2 \mid X_2=2]=\sum_{u=1}^{n^2}\sum_{\{i,j\ :\ i\cdot j=u\}}uP(X_1=i\ \cap\ X_2=j\mid X_1=2)=$$ $$=\sum_{u=1}^{n^2}\sum_{\{i,j\ :\ i\cdot j=u\}}u\frac{P(X_1=i\ \cap\ X_2=j\ \cap X_1=2)}{P(X_1=2)}=$$ $$=\sum_{u=1}^{n^2}\sum_{\{j\ :\ 2\cdot j=u\}}u\frac{P(X_1=2\ \cap\ X_2=j)}{P(X_1=2)}=$$ $$=\sum_{j=1}^n2jP(X_2=j\mid X_1=2)=E[2X_2\mid X_1=2].$$
$$E[X\mid X=x]=\sum_{i=1}^niP(X=i\mid X=x)=x.$$