Answer to Question #201630 in Statistics and Probability for miaz

Question

a continuous random variable T has the following probability density function.

f(t) = { 32t²/128, -4≤t<4

0, otherwise

a. find E(t)

b. find Var(T)

c. Var(2T+3)

Answer 1

$f(t)=\begin{cases} 32t^2/128 &\text{if } -4\leq t<4 \\ 0 &\text{otherwise } \end{cases}$

First of all we have to check whether this function is probability density function

So for pdf $\int_{-\infty}^\infty f(t)dx=1$

$\int_{-\infty}^\infty f(t)dt=\int_{-4}^4 \dfrac{32t^2}{128}dt$

$= \dfrac{32}{128}\times\dfrac{1}{3}[t^3]_{-4}^4\\\ \\= \dfrac{32}{128}\times\dfrac{1}{3}[4^3-(-4)^3]\\\ \\=\dfrac{32}{128}\times \dfrac{1}{3}\times128\\\ \\=\dfrac{32}{3}\neq 1$

So, it is not a probability density function.

(a) $E(t)=\int_{-\infty}^\infty tf(t)dt=\int_{-4}^4 \dfrac{32t^3}{128}dt=\dfrac{32}{128}\times\dfrac{1}{4}[t^4]_{-4}^4=\dfrac{8}{128}[256-256]=0$

Hence, $E(t)=0$

(b) $Var(t)$ $=E[(t-\mu)^2]= \int_{-\infty}^\infty (t-0)^2f(t)dt= \int_{-\infty}^\infty t^2f(t)dt=\int_{-4}^4 \dfrac{32}{128}t^4$

$\implies \dfrac{32}{128}\times \dfrac{1}{5}[t^5]_{-4}^4=\dfrac{32}{640}[1024-(-1024)]=\dfrac{32\times 2048}{640}=102.4$

(c) $Var(2t+3)=E[(2t+3-\mu)^2]= \int_{-\infty}^\infty (2t+3-0)^2f(t)dt= \int_{-\infty}^\infty (2t+3)^2f(t)dt$

$\implies \int_{-\infty}^\infty (4t^2+12t+9)f(t)dt=4 \int_{-\infty}^\infty t^2f(t)dt+12 \int_{-\infty}^\infty tf(t)dt+9 \int_{-\infty}^\infty f(t)dt$

We have ,

$\int_{-\infty}^\infty f(t)dt=32/3\\\ \\ \int_{-\infty}^\infty tf(t)dt=0\\\ \\ \int_{-\infty}^\infty t^2f(t)dt=102.4$

So,

$\implies \int_{-\infty}^\infty (4t^2+12t+9)f(t)dt=4 \int_{-\infty}^\infty t^2f(t)dt+12 \int_{-\infty}^\infty tf(t)dt+9 \int_{-\infty}^\infty f(t)dt\\\ \\\implies 4(102.4)+12\cdot(0)+9\times\dfrac{32}{3}=409.6+0+96=505.6$