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Q1. a)
X={1μlog(2u),u∈(0,12]−1μlog(2−2u),u∈(12,1),μ>0,μ∈R,u∼Uniform(0,1)
i) Let's first find the CDF for X:
- Case 1: u∈(0,12]⇒X∈(−∞,0]
FX(x)=P(X≤x)=P(1μlog(2u)≤x)=P(log(2u)≤μx)=P(elog(2u)≤eμx)=
=P(2u≤eμx)=P(u≤eμx2)=FU(eμx2)
We have u∼Uniform(0,1)⇒FU(u)=u
⇒FX(x)=FU(eμx2)=eμx2, when u∈(0,12],x∈(−∞,0]
- Case 2: u∈(12,1)⇒X∈(0,+∞)
FX(x)=P(X≤x)=P(−1μlog(2−2u)≤x)=P(log(2−2u)≥−μx)=
=P(elog(2−2u)≥e−μx)=P(2−2u≥e−μx)=P(u≤1−e−μx2)=
=FU(1−e−μx2)=1−e−μx2
⇒FX(x)={eμx2,when u∈(0,12],x∈(−∞,0]1−e−μx2,when u∈(12,1),x∈(0,+∞)
Now, let's compute the PDF for X. At values of x where FX is differentiable,
Plain Text
Q1. a)
$$X = \begin{cases} \frac{1}{\mu} \log(2u), & u \in (0, \frac{1}{2}] \\ -\frac{1}{\mu} \log(2-2u), & u \in (\frac{1}{2}, 1) \end{cases}, \mu > 0, \mu \in \mathbb{R}, u \sim \text{Uniform}(0,1)$$
i) Let's first find the CDF for $X$:
- **Case 1:** $u \in (0, \frac{1}{2}] \Rightarrow X \in (-\infty, 0]$
$$F_X(x) = P(X \leq x) = P\left(\frac{1}{\mu} \log(2u) \leq x\right) = P\left(\log(2u) \leq \mu x\right) = P\left(e^{\log(2u)} \leq e^{\mu x}\right) =$$
$$= P(2u \leq e^{\mu x}) = P(u \leq \frac{e^{\mu x}}{2}) = F_U\left(\frac{e^{\mu x}}{2}\right)$$
We have $u \sim \text{Uniform}(0,1) \Rightarrow F_U(u) = u$
$$\Rightarrow F_X(x) = F_U\left(\frac{e^{\mu x}}{2}\right) = \frac{e^{\mu x}}{2}, \text{ when } u \in (0, \frac{1}{2}], x \in (-\infty, 0]$$
- **Case 2:** $u \in (\frac{1}{2}, 1) \Rightarrow X \in (0, +\infty)$
$$F_X(x) = P(X \leq x) = P(-\frac{1}{\mu} \log(2-2u) \leq x) = P(\log(2-2u) \geq -\mu x) =$$
$$= P(e^{\log(2-2u)} \geq e^{-\mu x}) = P(2-2u \geq e^{-\mu x}) = P(u \leq 1 - \frac{e^{-\mu x}}{2}) =$$
$$= F_U(1 - \frac{e^{-\mu x}}{2}) = 1 - \frac{e^{-\mu x}}{2}$$
$$\Rightarrow F_X(x) = \begin{cases} \frac{e^{\mu x}}{2}, & \text{when } u \in (0, \frac{1}{2}], x \in (-\infty, 0] \\ 1 - \frac{e^{-\mu x}}{2}, & \text{when } u \in (\frac{1}{2}, 1), x \in (0, +\infty) \end{cases}$$
Now, let's compute the PDF for $X$. At values of $x$ where $F_X$ is differentiable,