Questions
I have the following problems on my probability module:
Qu1
Suppose that X is uniformly distributed over $[−3, 3]$.
(a) Find the probability that the first digit of $X$ after the decimal point is $3$.
(b) Find the distribution function of the random variable $|X|$ and use it to identify the distribution of $|X|$.
(c) Find the density function of the random variable $1/|X|$.
Qu2
Let $X$ be a standard exponential random variable, meaning that $X ∼ Exp(1)$. Prove that the random variable $Y = \lceil X \rceil$ is $Geom(\lambda)$ and find $\lambda$.
My answers and ideas
Qu1
(a) $\mathbb P($First digit after deciaml point of X is 3$)$$=\mathbb P(-2.4\lt X\lt -2.2)+\mathbb P(-1.4\lt X\lt -1.2)+\mathbb P(-0.4\lt X\lt -0.2)+\mathbb P(0.2\lt X\lt 0.4)+\mathbb P(1.2\lt X\lt 1.4)+\mathbb P(2.2 \lt X\lt 2.4)=6\mathbb P(0.2\lt X\lt 0.4)=6\frac{0.2}{3-(-3)}=\frac{1}{5}$
Im not 100% sure this is correct be great if someone could check my workings :)
(b) $|X| \Rightarrow$ $f(x)=\begin{cases} \frac{1}{3-0}, & \text{if $0\le x \le 3$} \\ 0, & \text{o/w} \end{cases}$
$\Rightarrow F(x)=\int_{-\infty}^{x}f(t)dt=\int_{-\infty}^{x}\frac{1}{3}dt=\frac{1}{3}(x-0)=\frac{x}{3}$
Even less sure with this one :/
(c) Not confident tackling as I'm unhappy with (b)
Qu2
I am clueless on how to approach this, tried drawing various distribution graphs. But they havent seemed to help me :(
Be great if anyone could help me by checking solutions and/or helping me with the questions I cannot do :)
For future reference, it is recommended that you post your questions separately, so Question 1 and Question 2 should be different posts.
One way to conceptualize part (a) is to observe that $Y = 10(X+3)$ is uniformly distributed on $[0, 60]$, and if the the tenths digit, which is first digit after the decimal point of $X$, is $3$, that means the units digit of $Y$ is $3$. This means the set of desired outcomes is $$Y \in [3, 4) \cup [13, 14), \cup \cdots \cup [53, 54).$$ Since this comprises six intervals of unit length, the total length is $6$ out of a total length of $60$, hence the desired probability is $1/10$.
For the second part, you need to rely less on general formulas and reason from first principles. You cannot blindly integrate the density in the way you have done.
Instead, note that the support of $X$ is on $[-3,3]$, so if $W = |X|$, the support of $W$ is on $[0,3]$. So we can already establish $\Pr[W < 0] = 0$ and $\Pr[W \le 3] = 1$. Then for any $w \ge 0$, we must have $$\Pr[W \le w] = \Pr[|X| \le w] = \Pr[-w \le X \le w] = \Pr[X \le w] - \Pr[X < -w].$$ From here, we can take two approaches. One is to differentiate with respect to $w$, giving $$f_W(w) = f_X(w) - (-f_X(-w)) = 2f_X(w) = 2 \cdot \frac{1}{6} \mathbb 1 (0 \le w \le 3) = \frac{1}{3} \mathbb 1 (0 \le w \le 3),$$ or we can directly substitute the CDF of $X$, which is $$\Pr[X \le x] = \begin{cases}0, & x < -3 \\ (x+3)/6, & -3 \le x \le 3 \\ 1, & x > 3, \end{cases}$$ to obtain $$\begin{align} \Pr[W \le w] &= \begin{cases} 0, & w < 0 \\ (w+3)/6 - (-w+3)/6, & 0 \le w \le 3 \\ 1, & w > 3 \end{cases} \\ &= \begin{cases} 0, & w < 0 \\ w/3, & 0 \le w \le 3 \\ 1, & w > 3. \end{cases} \end{align}$$
For part (c), the random variable $1/|X|$ is simply $1/W$ from the earlier part. Note that the support of $1/W$ is on $[1/3, \infty)$. Thus compute $$\Pr[1/W \le w] = \Pr[W \le 1/w]$$ for some $w \in [1/3, \infty)$.
I addressed Question 2 in a different answer here.