Conditioned Probability Task

Question

I have this task with two questions I would like to compute in Mathematica.

Suppose a pair of random variables $(X, Y)$ has a equal distribution on the following 7 points:

$\begin{matrix} & x & y\\ 1 & -1 & 0\\ 2 & 0 & 0\\ 3 & 1 & 0\\ 4 & -2& 1\\ 5 & 2 & 1\\ 6 & -1 & 3\\ 7 & 1 & 3 \end{matrix}$

The simultaneously probability function $p(x, y)$ is then given by: $p (1,0) = \ldots = P (1, 3) = 1/7 \text{ and } p(x, y) = 0$ for all other points

1) Find the conditional mean $E (Y\; |\; X = x)$ and the conditional variance $V (Y\; | \;X = x)$ for $x = -2, -1,0,1,2.$

2) Compute the variance on Y and the variance on the conditional mean of Y given X

This additional information might help you:

E(X)=0
E(Y)=8/7

I have tried over and over but without any luck.

Ps. I use Mathematica 8.

Answer 1

Assuming you have version 9 you can do the following.

data = {{-1, 0}, {0, 0}, {1, 0}, {-2, 1}, {2, 1}, {-1, 3}, {1, 3}};

dist = EmpiricalDistribution[data];

Table[Expectation[y \[Conditioned] x == i, {x, y} \[Distributed] dist], {i, -2, 2}]

(*{1, 3/2, 0, 3/2, 1}*)

Note: Conditional probabilities and expectations didn't work for EmpiricalDistribution in version 8. In that case you could code this up yourself as...

Table[Mean[Cases[data, {i, y_} :> y]], {i, -2, 2}]

(* {1, 3/2, 0, 3/2, 1} *)

Edit: An incredibly inefficient but distribution-based solution in M8 is to use ProbabilityDistribution.

pdist = 
 ProbabilityDistribution[Block[{tally = Tally[data], nprobs},
   nprobs = Normalize[tally[[All, 2]], Total];
   Piecewise[
    Transpose[{nprobs, (And @@ Thread[{x, y} == #]) & /@ 
       tally[[All, 1]]}]]
   ], {x, -3, 3, 1}, {y, -3, 3, 1}]

Table[
 Expectation[
  y \[Conditioned] x == i, {x, y} \[Distributed] pdist], {i, -2, 2}]

(* {1, 3/2, 0, 3/2, 1} *)

For completeness, to compute the variance using Expectation it gets a little messy.

mu = Expectation[y \[Conditioned] x == i, {x, y} \[Distributed] pdist];
Table[Expectation[(y - mu)^2 \[Conditioned] 
   x == i, {x, y} \[Distributed] pdist], {i, -2, 2}]

(*{0, 9/4, 0, 9/4, 0}*)

Answer 2

I should perhaps make this post a comment and not an answer. However, I wish to fully support the comments of @AndyRoss (and have +1 his answer).

cas = Cases[list, {#, y_} :> y] & /@ Range[-2, 2];
ans = {Mean[#], Mean[(# - Mean[#])^2]} & /@ cas;
Style[Prepend[
   MapThread[Prepend[#1, #2] &, {ans, Range[-2, 2]}], {"x", 
    "E[Y|X=x]", "Var[Y|X=x]"}] // Grid, 20]

I wished to make the following points (already made):

1. The underlying discrete uniform distribution make all the answers a matter of counting and the Mathematica functions are not really necessary.
2. Variance uses unbiased estimator and will (appropriately) fail for lists with one element. Using it in this context for larger lists lead to answer 9/2 due to correction factor ($n-1$ rather than $n$), so the definition of variance provides the suitable answer. (noting the cases: {{1}, {0, 3}, {0}, {0, 3}, {1}})
3. ProbabilityDistribution is a nice function but only complicates and as @AndyRoss has stated is inefficient for this specific problem. (This is really a restatement of 1).