Question

1)-Let X be a random variable with density function as follows:

2(x − 1) , 1 < x < 2 f(t) =
0 , elsewhere
• Find the expectation of x, and the variance.
• If g(x) = x2 + x − 2, where x has the same above density function. 1. E(g(x)).
2. E(2)

Answer 1

Answer on Question #49889-Math-Statistics and Probability

1) Let $X$ be a random variable with density function as follows:

$$f(x) = \begin{cases} 2(x - 1), & 1 < x < 2 \\ 0, & \text{elsewhere} \end{cases}$$

Find the expectation of $x$, and the variance.

If $g(x) = x^2 + x - 2$, where $x$ has the same above density function.

1. $E(g(x))$.

2. $E(2)$.

Solution

The expectation of $X$ is

$$E(X) = \int_{1}^{2} x \cdot 2(x - 1) \, dx = 2 \int_{1}^{2} (x^2 - x) \, dx = 2 \left(\frac{x^3}{3} - \frac{x^2}{2}\right)_{1}^{2} = \frac{5}{3}.$$

The variance of $X$ is

$$\operatorname{Var}(X) = E(X^2) - E(X)^2.$$ $$E(X^2) = \int_{1}^{2} x^2 \cdot 2(x - 1) \, dx = 2 \int_{1}^{2} (x^3 - x^2) \, dx = 2 \left(\frac{x^4}{4} - \frac{x^3}{3}\right)_{1}^{2} = \frac{17}{6}.$$ $$\operatorname{Var}(X) = \frac{17}{6} - \left(\frac{5}{3}\right)^{2} = \frac{1}{18}.$$

If $g(x) = x^2 + x - 2$, where $x$ has the same above density function.

1.

$$E(g(x)) = \int_{1}^{2} (x^2 + x - 2) \cdot 2(x - 1) \, dx = 2 \int_{1}^{2} (x^3 - 3x + 2) \, dx = 2 \left(\frac{x^4}{4} - 3\frac{x^2}{2} + 2x\right)_{1}^{2} = \frac{5}{2}.$$

2.

$$E(2) = \int_{1}^{2} 2 \cdot 2(x - 1) \, dx = 4 \left(\frac{x^2}{2} - x\right)_{1}^{2} = 2.$$

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