Grinstead-Snell-Probability · Problem 6.3.2

Sums of Random Variables

Mathematics · expected-value-variance-continuous

Problem

Let XX be a random variable with range [−1,1][-1, 1] and fXf_X its density function. Find μ(X)\mu(X) and σ2(X)\sigma^2(X) if, for ∣x∣>1|x| > 1, fX(x)=0f_X(x) = 0, and for ∣x∣<1|x| < 1, (a) fX(x)=(3/4)(1−x2)f_X(x) = (3/4)(1 - x^2). (b) fX(x)=(π/4)cos⁡(πx/2)f_X(x) = (\pi/4)\cos(\pi x/2). (c) fX(x)=(x+1)/2f_X(x) = (x+1)/2. (d) fX(x)=(3/8)(x+1)2f_X(x) = (3/8)(x+1)^2.
Intuitive explanation

The mean μ(X)=∫−11x⋅fX(x) dx\mu(X) = \int_{-1}^{1} x \cdot f_X(x) \, dx captures the center of mass of the distribution, while variance σ2(X)=∫−11x2⋅fX(x) dx−μ(X)2\sigma^2(X) = \int_{-1}^{1} x^2 \cdot f_X(x) \, dx - \mu(X)^2 measures spread around that center. For each density, you'll compute these integrals directly—watch for symmetry to simplify: if fXf_X is even, the mean vanishes immediately, saving work. The key is careful polynomial or trigonometric integration, then combining the results into the variance formula.

Intuitive explanation

The mean μ(X)=∫−11x⋅fX(x) dx\mu(X) = \int_{-1}^{1} x \cdot f_X(x) \, dx captures the center of mass of the distribution, while variance σ2(X)=∫−11x2⋅fX(x) dx−μ(X)2\sigma^2(X) = \int_{-1}^{1} x^2 \cdot f_X(x) \, dx - \mu(X)^2 measures spread around that center. For each density, you'll compute these integrals directly—watch for symmetry to simplify: if fXf_X is even, the mean vanishes immediately, saving work. The key is careful polynomial or trigonometric integration, then combining the results into the variance formula.

Related theory

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