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An exponential random variable t has a unit mean E{t}=1. What is the probability density function (pdf) of the random variable t?

A) f(t)=e⁻ᵗ
B) f(t)=eᵗ
C) f(t)=1−e⁻ᵗ
D) 2f(t)=t²

User Kantholy
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1 Answer

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Final answer:

The probability density function for an exponential random variable t with a unit mean is f(t) = e^(-t). Option A) f(t) = e^(-t) is the correct answer.

Step-by-step explanation:

The question asks for the probability density function (pdf) of an exponential random variable with a unit mean (E{t}=1). In an exponential distribution, the mean μ is the reciprocal of the decay parameter m (μ = 1∕m). Since μ = 1 for this problem, m would also equal 1, which gives us the pdf formula f(t) = me-mt, where t ≥ 0.

Subsequently, we plug in m = 1 to get f(t) = e-t. From the given options, the correct probability density function is A) f(t) = e-t which matches our derived formula.

User Mathias Schrooten
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