Conditional variance and total variance

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SUMMARY

The discussion rigorously establishes the law of total variance: Var(X) = E[Var(X|Y)] + Var(E[X|Y]). It clarifies that Var(X|Y) is a random variable dependent on Y, and the total variance of X decomposes into the expected conditional variance plus the variance of the conditional expectation. The example with a mixture of two normal distributions N(μ1,σ²) and N(μ2,σ²) demonstrates that the total variance exceeds the average of the conditional variances due to the variability in the conditional means. The discussion definitively confirms that the variability of the conditional means is not accounted for within E[Var(X|Y)] and must be included separately as Var(E[X|Y]).

PREREQUISITES

  • Conditional expectation and conditional variance concepts
  • Law of total variance and its mathematical proof
  • Probability distributions, specifically normal distributions N(μ,σ²)
  • Basic understanding of random variables and expectation operators E[·]

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  • Study the law of total expectation and its applications
  • Explore mixture models and their variance decomposition
  • Learn advanced conditional probability techniques in measure-theoretic probability
  • Apply variance decomposition in practical statistical modeling and inference

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Statisticians, data scientists, probabilists, and researchers working with hierarchical models, mixture distributions, or conditional probability frameworks will benefit from understanding the precise decomposition of variance and the role of conditional expectations in variance analysis.

Kakashi
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Variance is the average of the squared distance from the mean of the distribution fX∣Y. When y is not specified, Var(X∣Y) is a random variable that depends on Y.

$$ Var(X|Y)=E[X^{2}|Y]-(E[X|Y])^{2} $$
Taking expectations:
$$ E[Var(X|Y)]=E[E[X^{2}|Y]-(E[X|Y])^{2}]=E[E[X^{2}|Y]]-E[(E[X|Y])^{2}]=E[X^{2}]-E[(E[X|Y])^{2}] $$
and
$$ Var(E[X|Y])=E[(E[X|Y])^{2}]-(E[E[X|Y]])^{2} $$

Adding these gives:

$$ E[Var(X|Y)]+Var(E[X|Y])=E[X^{2}]-(E[X])^{2}=Var(X) $$

Thinking from a more practical point of view: when we divide the real line into small intervals, run the experiment many times, count how often X falls within each interval, and then divide by both the total number of trials and the length of the interval, we obtain fX, from which we can compute E[X] and Var(X).


How do we know that there is another random variable Y that we can condition on, find fX∣Y, and compute E[X|Y] and Var(X|Y) this way? It seems like we would have to repeat this for many values of y, which looks like more work than working directly with X.

Why isn’t the variability in X just the weighted average of the variances of the different conditional distributions fX∣Y or PX|Y?

Why do we need to include the variability of the mean, Var(E[X∣Y])?

Isn’t the variability of the mean already somehow accounted for in E[Var(X∣Y)], since each conditional variance is computed around its own mean?
 
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To simplify, imagine X is drawn randomly with 50% probability from either of 2 distributions N(μ12) and N(μ22), with μ1 - μ2 >> σ.
Do you imagine the variance of X will be equal to the mean of the two separate variances, i.e. σ2? It will obviously depend on the difference between the means as well.
Kakashi said:
Isn’t the variability of the mean already somehow accounted for in E[Var(X∣Y)], since each conditional variance is computed around its own mean?
No. For that very reason, it isn't accounted for.
 
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