Continuous Random Variables: PDF, Uniform & Exponential
“PDF integration, CDF properties, Continuous Uniform, Exponential distribution, and the memoryless property.”
1.Continuous Probability Density & The Memoryless Exponential
§ 2.1.For continuous random variables, probabilities correspond to intervals. The Probability Density Function (PDF) f(x) represents density, not point probability: P(X in [x, x + dx]) ≈ f(x) dx.
Continuous Uniform Distribution U(a, b):
Every subinterval of equal length within [a, b] has identical probability. f(x) = 1 / (b - a).
- E[X] = (a + b) / 2
- Var(X) = (b - a)^2 / 12
Exponential Distribution Exp(lambda):
Models the continuous waiting time until the first occurrence of a Poisson event.
f(x) = lambda e^{-lambda x} for x >= 0.
F(x) = 1 - e^{-lambda x}.
Proof of the Memoryless Property:
P(X > s + t | X > s) = P(X > s + t ∩ X > s) / P(X > s) = P(X > s + t) / P(X > s)
= e^{-lambda (s + t)} / e^{-lambda s} = e^{-lambda t} = P(X > t).
2. Worked Practice Problem: Memoryless Exponential Survival
Do not compute complex conditional integration; apply the memoryless property directly.
Problem: The lifetime of a server GPU (in thousands of hours) follows an Exponential distribution with rate parameter lambda = 0.5. Given that a GPU has already operated without failure for 2,000 hours (X > 2), compute the probability that it will survive for at least 5,000 hours in total (X > 5).
Step 1: Identify the conditional probability required:
P(X > 5 | X > 2)
Step 2: Express in the memoryless form P(X > s + t | X > s):
Here s = 2 and s + t = 5 => t = 5 - 2 = 3.
Step 3: Apply the memoryless property:
P(X > 2 + 3 | X > 2) = P(X > 3)
Step 4: Compute survival probability for Exponential distribution:
P(X > t) = 1 - F(t) = e^{-lambda * t}
Substitute lambda = 0.5 and t = 3:
P(X > 3) = e^{-(0.5 * 3)} = e^{-1.5}
Step 5: Calculate numerical value:
e^{-1.5} ≈ 0.22313 (approx 22.31%).
Conclusion: The probability of surviving an additional 3,000 hours given it survived 2,000 hours is e^{-1.5} ≈ 0.2231.
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