Final answer:
The mean (μ) and standard deviation (σ) for an exponential random variable are both equal to 1/m. To prove this, we use the moment generating function to find the first and second derivatives evaluated at t=0, indicating that μ = M'(0) = 1/m and σ = √(1/m²) =1/m, showing μ = σ.
Step-by-step explanation:
To show that the mean (μ) is equal to the standard deviation (σ) for an exponential random variable, consider the random variable X with an exponential distribution, denoted as X~Exp(m). The mean and the standard deviation of an exponential distribution are both given by the formula 1/m. The probability density function (pdf) of X is f(x) = me-mx, for x ≥ 0.
The moment generating function (MGF) for X is defined as M(t) = E(etx). To find the MGF, we integrate the product of etx and the pdf f(x) over all possible values of x. Hence, M(t) = ∫0∞ etxme-mx dx, which simplifies to M(t) = m/(m - t), for t < m. To find the mean, we differentiate M(t) with respect to t and evaluate at t = 0, yielding μ = M'(0) = 1/m.
For the variance, σ2, we differentiate the MGF twice and again evaluate at t = 0. This gives σ2 = M''(0) = 2/m2 - 1/m2 = 1/m2. Taking the square root of the variance, we obtain σ = √(1/m2) = 1/m, demonstrating that the standard deviation is indeed equal to the mean for an exponential random variable. Consequently, μ = σ.
The complete question is: Using the moment generating function for an exponential random variable, demonstrate that the mean (μ) is equal to the standard deviation (σ), which means μ=σ. is: