Processing math: 100%

NanoNets OCR for Handwritten Notes

Prompt:

Extract the text from the above document as if you were reading it naturally. 

Return the tables in html format.

Return the equations in LaTeX representation.

If there is an image in the document and image caption is not present, add a small description of the image inside the <img></img> tag; otherwise, add the image caption inside <img></img>.

Watermarks should be wrapped in brackets. Ex: <watermark>OFFICIAL COPY</watermark>. 

Page numbers should be wrapped in brackets. Ex: <page_number>14</page_number> or <page_number>9/22</page_number>. 

Prefer using  and  for check boxes.

Input:

Example conversation
Figure 1. Input image

Result:

Rendered

Q1. a)

X={1μlog(2u),u(0,12]1μlog(22u),u(12,1),μ>0,μR,uUniform(0,1)

i) Let's first find the CDF for X:

  • Case 1: u(0,12]X(,0]
FX(x)=P(Xx)=P(1μlog(2u)x)=P(log(2u)μx)=P(elog(2u)eμx)=
=P(2ueμx)=P(ueμx2)=FU(eμx2)

We have uUniform(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(Xx)=P(1μlog(22u)x)=P(log(22u)μx)=
=P(elog(22u)eμx)=P(22ueμx)=P(u1eμx2)=
=FU(1eμx2)=1eμx2
FX(x)={eμx2,when u(0,12],x(,0]1eμx2,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,

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