Conditional expectation for $2X+3 \mid Y$

Question

Given the pdf: $$f_{X,Y}(x,y)= \begin{cases} x+y,& 0 \le x \le 1, 0 \le y \le 1 \\0, & \text{otherwise}\end{cases}$$ Find $E(2X + 3\mid Y )$.

Using the formula for conditional expectation: $\int{xf(x|y)dx}$, where $x=2X+3$, so we have:

$$\int\limits_0^1{(2x+3)\frac{f(x,y)}{f_y(y)}dx}$$

Is this correct or do we have to do something different regarding the $2X+3$? Thanks.

Answer 1

Your approach is correct.

You can verify your result with Mathematica:

In[16]:= Expectation[
  2 X + 3 \[Conditioned] Y == y, {X, Y} \[Distributed] 
   ProbabilityDistribution[x + y, {x, 0, 1}, {y, 0, 1}]] // Normal

Out[16]= (13 + 24 y)/(3 (1 + 2 y))

And the confirmation that the result is correct, using the law of total expectation:

In[21]:= Expectation[(13 + 24 y)/(3 (1 + 2 y)), 
  y \[Distributed] 
   MarginalDistribution[
    ProbabilityDistribution[x + y, {x, 0, 1}, {y, 0, 1}], 2]] == 
 Expectation[
  2 x + 3, {x, y} \[Distributed] 
   ProbabilityDistribution[x + y, {x, 0, 1}, {y, 0, 1}]]

Out[21]= True