Question

Let U be a continuous random variable with the following probability density function: g(t) = 1+t -1<t< 0 1-t 0<t<1 0 otherwise a. Verify that g(t) is indeed a probability density function. [5] b. Compute the expected value, E(U), and variance, V(U), of U. (10)

Answer 1

a.

To verify that g(t) is indeed a probability density function, we need to check for two conditions:

Condition 1: g(t) is non-negative for all values of t in the support of U.

Condition 2: The integral of g(t) integrated over the entire support of U integrates to 1.

Condition 1

Case I: When -1 ≤ t ≤ 0:

=> -1 ≤ t

=> 0 ≤ 1 + t

=> 0 ≤ g(t)

=> g(t) is non-negative when -1 ≤ t ≤ 0

Case II: When 0 ≤ t ≤ 1:

=> t ≤ 1

=> 0 ≤ 1 - t

=> 0 ≤ g(t)

=> g(t) is non-negative when 0 ≤ t ≤ 1

Thus, g(t) is non-negative for all values of t in the support of U.

Condition 2

We find:

$\begin{align*} \int_{-1}^{1} g(t) \ dt &= \int_{-1}^{0} g(t) \ dt + \int_{0}^{1} g(t) \ dt \\ &= \int_{-1}^{0} (1+t) \ dt + \int_{0}^{1} (1-t) \ dt \\ &= \left[t+\frac{t^2}{2} \right ]_{t=-1}^{t=0} + \left[t - \frac{t^2}{2} \right ]_{t=0}^{t=1} \\ &= \left[\left(0+\frac{0^2}{2} \right ) - \left((-1) + \frac{(-1)^2}{2} \right )\right ] + \left[\left(1-\frac{1^2}{2} \right ) - \left(0-\frac{0^2}{2} \right ) \right ] \\ &= \left[\left(0+0 \right ) - \left(-1 + \frac{1}{2} \right )\right ] + \left[\left(1-\frac{1}{2} \right ) - \left(0-0 \right ) \right ] \\ &= \left[0 - \left(- \frac{1}{2} \right )\right ] + \left[\frac{1}{2}-0 \right ] \\ &= \frac{1}{2} + \frac{1}{2} \\ &= 1 \end{align*}$

Thus, the integral of g(t) integrated over the entire support of U equals to 1.

Thus, g(t) is a valid probability density function.

b.

The expected value of U is given by:

$\begin{align*} \bf E(U) &= \int_{-1}^{1} t*g(t) \ dt \\ &= \int_{-1}^{0} t*g(t) \ dt + \int_{0}^{1} t*g(t) \ dt \\ &= \int_{-1}^{0} t(1+t) \ dt + \int_{0}^{1} t(1-t) \ dt \\ &= \int_{-1}^{0} (t+t^2) \ dt + \int_{0}^{1} (t-t^2) \ dt \\ &= \left[\frac{t^2}{2}+\frac{t^3}{3} \right ]_{t=-1}^{t=0} + \left[\frac{t^2}{2} - \frac{t^3}{3} \right ]_{t=0}^{t=1} \\ &= \left[\left(\frac{0^2}{2}+\frac{0^3}{3} \right ) - \left(\frac{(-1)^2}{2}+\frac{(-1)^3}{3} \right ) \right ] + \left[\left(\frac{1^2}{2} - \frac{1^3}{3} \right ) - \left(\frac{0^2}{2} - \frac{0^3}{3} \right ) \right ] \\ &= \left[(0+0 ) - \left(\frac{1}{2}-\frac{1}{3} \right ) \right ] + \left[\left(\frac{1}{2} - \frac{1}{3} \right ) - (0-0) \right ] \\ &= \left[0 - \frac{3-2}{6} \right ] + \left[\frac{3-2}{6} -0 \right ] \\ &= -\frac{1}{6} + \frac{1}{6} \\ &= \bf 0 \ \ \ \ \ \ \ \ \ \ \ \ [ANSWER] \end{align*}$

Now, before finding the variance of U, we find:

$\begin{align*} E(U^2) &= \int_{-1}^{1} t^2*g(t) \ dt \\ &= \int_{-1}^{0} t^2*g(t) \ dt + \int_{0}^{1} t^2*g(t) \ dt \\ &= \int_{-1}^{0} t^2(1+t) \ dt + \int_{0}^{1} t^2(1-t) \ dt \\ &= \int_{-1}^{0} (t^2+t^3) \ dt + \int_{0}^{1} (t^2-t^3) \ dt \\ &= \left[\frac{t^3}{3}+\frac{t^4}{4} \right ]_{t=-1}^{t=0} + \left[\frac{t^3}{3} - \frac{t^4}{4} \right ]_{t=0}^{t=1} \\ &= \left[\left(\frac{0^3}{3}+\frac{0^4}{4} \right ) - \left(\frac{(-1)^3}{3}+\frac{(-1)^4}{4} \right ) \right ] + \left[\left(\frac{1^3}{3} - \frac{1^4}{4} \right ) - \left(\frac{0^3}{3} - \frac{0^4}{4} \right ) \right ] \\ &= \left[(0+0 ) - \left(-\frac{1}{3}+\frac{1}{4} \right ) \right ] + \left[\left(\frac{1}{3} - \frac{1}{4} \right ) - (0-0) \right ] \\ &= \left[0 - \left(\frac{-4+3}{12} \right ) \right ] + \left[\frac{4-3}{12} -0 \right ] \\ &= \left[0 - \left(\frac{-1}{12} \right ) \right ] + \left[\frac{1}{12} -0 \right ] \\ &= \frac{1}{12} + \frac{1}{12} \\ &= \frac{2}{12} \\ &= \frac{1}{6} \end{align*}$

Thus, the variance of U is given by:

$\begin{align*} \bf V(U) &= E(U^2) - [E(U)]^2 \\ &= \frac{1}{6} - 0^2 \\ & \bf \frac{1}{6} \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ANSWER] \end{align*}$

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