Let g(x) be the density function of a continuous random variable \xi. Find the function value of the cumulative distribution function of \xi at 2 if g(x)=\left\{\begin{array}{ccc} 0 & ext{if } & x

Question

Let g(x) be the density function of a continuous random variable \xi. Find the function value of the cumulative distribution function of \xi at 2 if g(x)=\left\{\begin{array}{ccc} 0 & \text{if } & x<1 \\{1\over4} x & \text{if } & 1\le x\le 3\\0 & \text{if } & 3<x \\\end{array}\right..简答题

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BlackTom AI · Accepted Answer

We start by restating the problem: g(x) is the density of a continuous random variable ξ, given piecewise as 0 for x < 1, (1/4) x for 1 ≤ x ≤ 3, and 0 for x > 3. We are asked to find Fξ(2), the value of the cumulative distribution function at x = 2. First, recall that the CDF Fξ(x) = P(ξ ≤ x) is the integral of the density up to x: Fξ(x) = ∫_{-∞}^{x} g(t) dt. Since g(t) = 0 for t < 1, and 1 ≤ t ≤ 3 uses g(t) = (1/4) t, while t > 3 is 0, we only need to integrate from t = 1 up to t = x when x lies in [1, 3]. Here x = 2, which lies in that interval. Compute the integral: Fξ(2) = ∫_{1}^{2} (1/4) t dt. The antiderivative of (1/4) t with respect to t is (1/8) t^2. Evaluating from 1 to 2 gives (1/8)(2^2) − (1/8)(1^2) = (1/8)(4 − 1) = (1/8)·3 = 3/8. Thus, the cumulative probability up to x = 2 equals 3/8, which is 0.375. If desired, this can be interpreted as the probability that the random variable ξ takes a value in the interval [1, 2], since the density is positive only there and increases linearly with x in that range.

Let g(x) be the density function of a continuous random variable \xi. Find the function value of the cumulative distribution function of \xi at 2 if g(x)=\left\{\begin{array}{ccc} 0 & \text{if } & x<1 \\{1\over4} x & \text{if } & 1\le x\le 3\\0 & \text{if } & 3<x \\\end{array}\right..简答题

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