Question

Let fX (x)=1/3 X[U(x)-U(x-2]+1/3 δ(x-1). Find
The expected value E[X]. B. the variance σx^2.

Answer 1

Answer on Question #48726 – Math – Statistics and Probability

Let $f_{X}(x) = \frac{1}{3} x[U(x) - U(x - 2)] + \frac{1}{3} \delta(x - 1)$. Find

The expected value $E[X]$. B. the variance $\sigma x^2$.

Solution

$$\begin{aligned} \mathrm{E}[X] &= \int_{-\infty}^{\infty} x f_{X}(x) \, dx = \int_{-\infty}^{\infty} x \cdot \left( \frac{1}{3} x [U(x) - U(x - 2)] + \frac{1}{3} \delta(x - 1) \right) dx \\ &= \frac{1}{3} \int_{-\infty}^{\infty} x^{2} \cdot [U(x) - U(x - 2)] \, dx + \frac{1}{3} \int_{-\infty}^{\infty} x \cdot \delta(x - 1) \, dx. \end{aligned}$$ $$\int_{-\infty}^{\infty} x \cdot \delta(x - 1) \, dx = 1.$$ $$\int_{-\infty}^{\infty} x^{2} \cdot [U(x) - U(x - 2)] \, dx = \int_{0}^{2} x^{2} \, dx = \left( \frac{x^{3}}{3} \right)_{0}^{2} = \frac{8}{3}.$$

So, the expected value is

$$\mathrm{E}[X] = \frac{1}{3} \cdot \frac{8}{3} + \frac{1}{3} \cdot 1 = \frac{11}{9}.$$

The variance is

$$\sigma_{x}^{2} = \mathrm{E}[X^{2}] - (\mathrm{E}[X])^{2}, \text{ where}$$ $$\begin{aligned} \mathrm{E}[X^{2}] &= \int_{-\infty}^{\infty} x^{2} f_{X}(x) \, dx = \int_{-\infty}^{\infty} x^{2} \cdot \left( \frac{1}{3} x [U(x) - U(x - 2)] + \frac{1}{3} \delta(x - 1) \right) dx \\ &= \frac{1}{3} \int_{-\infty}^{\infty} x^{3} \cdot [U(x) - U(x - 2)] \, dx + \frac{1}{3} \int_{-\infty}^{\infty} x^{2} \cdot \delta(x - 1) \, dx. \end{aligned}$$ $$\int_{-\infty}^{\infty} x^{2} \cdot \delta(x - 1) \, dx = 1^{2} = 1.$$ $$\int_{-\infty}^{\infty} x^{3} \cdot [U(x) - U(x - 2)] \, dx = \int_{0}^{2} x^{3} \, dx = \left( \frac{x^{4}}{4} \right)_{0}^{2} = \frac{2^{4}}{4} = 4.$$

So, $\mathrm{E}[X^2] = \frac{1}{3} \cdot 4 + \frac{1}{3} \cdot 1 = \frac{5}{3}$.

Thus,

$$\sigma_{x}^{2} = \frac{5}{3} - \left( \frac{11}{9} \right)^{2} = \frac{5 \cdot 27 - 121}{81} = \frac{14}{81}$$

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