Show that the differential equation of Exercise 6.30 with b = c = 0 and σ > 0 yields a normal distribution.
Answer to relevant QuestionsIn the proof of Theorem 6.6 we twice differentiated the moment– generating function of the normal distribution with respect to t to show that E(X) = µ and var(X) = σ2. Differentiating twice more and using the formula of ...Show that if a random variable has a uniform density with the parameters α and β, the rth moment about the mean equals (a) 0 when r is odd; (b) 1 / r + 1 (β – α / 2)r when r is even. If X and Y have a bivariate normal distribution and U = X + Y and V = X – Y, Find an expression for the correlation coefficient of U and V. The mileage (in thousands of miles) that car owners get with a certain kind of radial tire is a random variable having an exponential distribution with θ = 40. Find the probabilities that one of these tires will last (a) ...If Z is a random variable having the standard nor–mal distribution, find the respective values z1, z2, z3, and z4 such that (a) P(0 < Z < z1) = 0.4306; (b) P(Z ≥ z2) = 0.7704; (c) P(Z > z3) = 0.2912; (d) P(–z4 ≤ Z ...
Post your question