Quadratic Equation of uniform random variable

Question

By looking at the mgf, The thing is got to know is that $X \sim U(0,1) $

But how do we find the value if $ \alpha \space \text{and} \space \beta$ .

Answer 1

$X \sim U[0,1]$, $$48X^2-40X+3> 0$$ $$(12X-1)(4X-3) > 0$$

That is $X< \frac1{12}$ or $X>\frac34$.

Hence $$\alpha = \frac1{12} + \left(1 - \frac34\right)=\frac{1+12-9}{12}=\frac13$$

Now let $Y =\ln X$.

$$Y^2+2Y-3>0$$ $$(Y+3)(Y-1) >0$$ That is $Y<-3$ or $Y>1$ which is equivalent to $X < e^{-3}$ or $X > e$. Hence,

$$\beta = e^{-3}$$ $$\ln \beta = -3$$

$$\alpha - 2 \ln \beta = \frac13 - 2(-3)=6\frac13=\frac{19}{3}$$

Answer 2

Hint:

You have found the distribution of $X$, so you need to

• solve the inequalities,
• which then gives you the probabilities as the proportion of $[0,1]$ where they are positive
• and thus the value of the expression.

The inequalities are essentially quadratic, and on $[0,1]$ the curves look like this (which you can use to check you have calculated correctly)