📘 Law of Total Probability with Conditional Distribution Equality

🔹 Goal

We want to understand why:

$P(X = a) = P(Y = a)$

when:

$P(X = a \mid Z = z) = P(Y = a \mid Z = z)$

🔹 Step 1: Start with Law of Total Probability

We begin with:

$P(X = a) = \sum_z P(X = a \mid Z = z),P(Z = z)$

👉 This means:

We calculate the probability of X = a by considering all possible values of Z

🔹 Step 2: Intuition Behind This Formula

Think of it as:

👉 Break the problem into cases based on Z
👉 Then combine (weight) each case by how likely Z is

So:

$P(X = a \mid Z = z)$ → behavior in each case
$P(Z = z)$ → importance (weight) of that case

🔹 Step 3: Use the Key Assumption

We are given:

$P(X = a \mid Z = z) = P(Y = a \mid Z = z)$

👉 Replace X with Y inside the summation:

$P(X = a) = \sum_z P(Y = a \mid Z = z),P(Z = z)$

🔹 Step 4: Recognize the Same Formula for Y

Now observe:

$P(Y = a) = \sum_z P(Y = a \mid Z = z),P(Z = z)$

👉 This is again the law of total probability, but applied to Y

🔹 Step 5: Final Conclusion

Since both expressions are identical:

$P(X = a) = P(Y = a)$

🔹 Plain English Interpretation

👉 If X and Y behave the same under every condition Z
👉 Then overall (unconditionally), they also behave the same

🔹 Real-Life Analogy

Let:

Z = type of customer
X = amount spent on product A
Y = amount spent on product B

If:

👉 For every customer type, spending patterns are identical

Then:

👉 Overall spending distribution is also identical

🔹 Key Insight

👉 Matching conditional distributions + same weights = same overall distribution

🔹 One-Line Summary

👉 If $P(X \mid Z) = P(Y \mid Z)$ , then $P(X) = P(Y)$

Additional menu