Prompt Title: Symmetric CRV Expectation Calculation.

To show that E(X) = µ, we can use the following steps:

1. Using the symmetry of the PDF, we have:

   E(X) = ∫_{-∞}^{∞} x fX(x) dx

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     = ∫_{-∞}^{0} x fX(x) dx + ∫_{0}^{∞} x fX(x) dx
   
     = ∫_{0}^{∞} (-x) fX(-x) dx + ∫_{0}^{∞} x fX(x) dx
   
     (substituting -x for x in the first integral)
   
     = 2 ∫_{0}^{∞} x fX(x) dx
   

2. Let Y = X - µ. Then Y is also a CRV with a symmetric PDF, since:

   fY(µ + y) = fX(µ + y - µ) = fX(y) = fX(-(y-µ)) = fY(µ - y)

   Therefore, using the definition of expectation and the substitution Y = X - µ, we have:

   E(X) = E(Y + µ) = E(Y) + µ

3. Since Y = X - µ is symmetric about 0, we have:

   ∫_{-∞}^{∞} Y fY(Y) dY = 0

   Substituting Y = X - µ and rearranging, we get:

   ∫_{-∞}^{∞} (X - µ) fX(X) dX = µ ∫_{-∞}^{∞} fX(X) dX

   Using the definition of expectation and the fact that fX(X) is a PDF (i.e., integrates to 1), we have:

   E(X) - µ = 0

4. Combining steps 1 and 3, we get:

   E(X) = 2 ∫_{0}^{∞} x fX(x) dx = 2 ∫_{0}^{∞} (x - µ) fX(x) dx + 2µ ∫_{0}^{∞} fX(x) dx

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    = 2 E(X - µ) + 2µ
   

   Substituting E(X - µ) = 0 (from step 3), we get:

   E(X) = 2µ

   Therefore, E(X) = µ, which completes the proof.