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How to calculate the expectation and variance of a complex probability distribution

Mathematica Asked on October 26, 2020

Assuming that the continuous random variables X1 and X2 are independent of each other, and the variances exist, the probability densities of X1 and X2 are $f_{1}(x)$
and $f_{2}(x)$, the probability density of random variable $Y_{1}$ is $f_{Y_{1}}(y)=frac{1}{2}left[f_{1}(y)+f_{2 }(y)right]$, random variablemathrm${Y}_{2}=
frac{1}{2}left(X_{1}+X_{2}right)$
. Which of the following statements is correct (the answer is D)?

$$begin{array}{c}
&(A)& E Y_{1}>E Y_{2}, D Y_{1}>D Y_{2}
&(B)& E Y_{1}=E Y_{2}, D Y_{1}=D Y_{2}
&(C)& E Y_{1}=E Y_{2}, D Y_{1}<D Y_{2}
&(D)& E Y_{1}=E Y_{2}, D Y_{1}>D Y_{2}
end{array}$$

When I use the normal distribution to verify the D option, the following code keeps running:

Y1 = ProbabilityDistribution[(1/
     2) (PDF[NormalDistribution[μ1, σ1], x] + 
     PDF[NormalDistribution[μ2, σ2], x]), {x, -Infinity, 
   Infinity}]


Expectation[Y1, Y1 [Distributed] Y1]
Variance[Y1]
Y2 = TransformedDistribution[
  1/2 (x1 + x2), {x1 [Distributed] 
    NormalDistribution[μ1, σ1], 
   x2 [Distributed] NormalDistribution[μ2, σ2]}]
Expectation[Y2, Y2 [Distributed] Y2]
Variance[Y2]

How can I improve the code to get the desired result?

One Answer

Clear["Global`*"]

In defining Y1 you need to specify the assumptions needed to make the distribution valid.

Y1 = ProbabilityDistribution[
   (1/2) (PDF[NormalDistribution[μ1, σ1], x] +
      PDF[NormalDistribution[μ2, σ2], x]),
   {x, -Infinity, Infinity},
   Assumptions -> {σ1 > 0, σ2 > 0}];

Mean[Y1]

(* (μ1 + μ2)/2 *)

which is equivalent to

Expectation[x, x [Distributed] Y1]

(* (μ1 + μ2)/2 *)

Note that you cannot use the distribution as a variable.

Variance[Y1]

(* 1/4 ((μ1 - μ2)^2 + 2 (σ1^2 + σ2^2)) *)

Y2 = TransformedDistribution[
  1/2 (x1 + x2), {x1 [Distributed] NormalDistribution[μ1, σ1], 
   x2 [Distributed] NormalDistribution[μ2, σ2]}]

(* NormalDistribution[μ1/2 + μ2/2, Sqrt[σ1^2/4 + σ2^2/4]] *)

Mean[Y2] // Simplify

(* (μ1 + μ2)/2 *)

Expectation[x, x [Distributed] Y2]

(* (μ1 + μ2)/2 *)

Variance[Y2] // Simplify

(* 1/4 (σ1^2 + σ2^2) *)

Correct answer by Bob Hanlon on October 26, 2020

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