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1 1 vote

Let $X$ be a continuous random variable.
$X$ has a constant PDF on $(0,1]$, and another constant PDF on $(1,2]$.
The PDF of $X$ is given by: $$ f_X(x)= \begin{cases} c, & \text{for } 0<x\leq 1,\\ d, & \text{for } 1<x\leq 2,\\ 0, & \text{otherwise}. \end{cases} $$ It is known that

$$ \mathbb{P}(1\leq X\leq 2) = 2\cdot\mathbb{P}(0\leq X<1)$$

What are the values of $c$ and $d$?

  1. $c=\frac{1}{2},\quad d=\frac{1}{2}$
     
  2. $c=\frac{1}{3},\quad d=\frac{2}{3}$
     
  3. $c=\frac{2}{3},\quad d=\frac{1}{3}$
     
  4. $c=\frac{1}{4},\quad d=\frac{3}{4}$

1 Answer

2 2 votes



Since the PDF is constant over each interval, the probability over each interval is just the height times the width:

  • $\mathbb{P}(0<X \leq 1)=c \cdot 1=c$
  • $\mathbb{P}(1<X \leq 2)=d \cdot 1=d$


From the condition, $d=2 c$.
Also, the total probability must be 1 , so:
$$
c+d=1
$$

Substituting $d=2 c$ gives:
$$
c+2 c=1 \Rightarrow 3 c=1 \Rightarrow c=\frac{1}{3}, \quad d=\frac{2}{3}
$$

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