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Verify pdf 2e^(−2x); P(1<x<3); mean

Short answer

The function is non-negative and ∫₀^∞ 2e^(−2x) dx = 1, so it's a valid pdf. P(1<X<3)=e^(−2)−e^(−6)≈0.1329. It is exponential with λ=2, thus mean=1/λ=0.5, variance=1/λ²=0.25.

If a random variable X has a function f(x) = 2e^(−2x) for x > 0; 0 for x ≤ 0. Find (i) Verify that the function is probability density function (ii) P(1 < x < 3) (iii) Find mean and variance

f(x) = 2e^(−2x) for x > 0 is an exponential pdf with λ = 2.

(i) Verify that f(x) is a pdf

  1. f(x) = 2e^(−2x) > 0 for all x > 0, and f(x) = 0 otherwise, so f(x) ≥ 0.
  2. ∫₀^∞ 2e^(−2x) dx = [−e^(−2x)]₀^∞ = 0 − (−1) = 1.

Both conditions hold, so f(x) is a valid probability density function.

(ii) P(1 < X < 3)

P(1 < X < 3) = ∫₁³ 2e^(−2x) dx = [−e^(−2x)]₁³ = e^(−2) − e^(−6)

= 0.1353 − 0.0025 = 0.1329

(iii) Mean and variance

For the exponential distribution with parameter λ:

  • Mean = 1/λ = 1/2 = 0.5
  • Variance = 1/λ² = 1/4 = 0.25

Check by integration: E(X) = ∫₀^∞ x·2e^(−2x) dx = 1/2; E(X²) = ∫₀^∞ x²·2e^(−2x) dx = 1/2; so Var = 1/2 − 1/4 = 1/4.

✓ Answer checked by Claude Sonnet 5.5

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