How to do part C (2 Viewers)

MonKey1919 · Sep 21, 2021

Pretty much what the title says. Never come across this type of question so I have no idea what to do.

Screen Shot 2021-09-21 at 2.19.59 pm.png

=)(= · Sep 21, 2021

It wants you to derive the cumulative distribution function

MonKey1919 · Sep 21, 2021

=)(= said:
It wants you to derive the cumulative distribution function

Nah CDF is given. F is the CDF so they want to PDF

=)(= · Sep 21, 2021

MonKey1919 said:
Nah CDF is given. F is the CDF so they want to PDF

Nah i mean like differentiate

=)(= · Sep 21, 2021

c

=)(= said:
Nah i mean like differentiate

cause you integrate to get the cdf

MonKey1919 · Sep 21, 2021

=)(= said:
c

cause you integrate to get the cdf

Ohhhh but there are two different equations. Do you diff them both and make it a piecewise function?

=)(= · Sep 21, 2021

I think so ye

=)(= · Sep 21, 2021

Damn that is a tough one

=)(= · Sep 21, 2021

yep do em in cases

CM_Tutor · Sep 21, 2021

If you think about tax scales questions from junior maths, the function we have here is similar, in that each piece has a constant added (the cumulative probability from earlier pieces. Hence

$\text{For $1 \le x < 3$,}\ F(x) = 0 + \cfrac{1}{4}(x-1)$

$\implies \text{as $x \to 3^-$}, F(x) \to \cfrac{1}{4}(3-1) = \cfrac{2}{4} = \cfrac{1}{2}$

$\text{So, at $x = 3$, it is necessary for $F(x) = \cfrac{1}{2}$, explaining the "$\cfrac{1}{2}$+" term in the next piece}$

$\text{For $3 \le x < 7$,}\ F(x) = \cfrac{1}{2} + \cfrac{1}{8}(x-3)$

$\implies \text{as $x \to 7^-$}, F(x) \to \cfrac{1}{2} + \cfrac{1}{8}(7-3) = \cfrac{1}{2} + \cfrac{4}{8} = 1$

$\text{So, for $x \ge 7$, it is necessary for $F(x) = 1$, as a CDF cannot exceed 1}$

=)(= · Sep 21, 2021

CM_Tutor said:
If you think about tax scales questions from junior maths, the function we have here is similar, in that each piece has a constant added (the cumulative probability from earlier pieces. Hence

$\text{For $1 \le x < 3$,}\ F(x) = 0 + \cfrac{1}{4}(x-1)$

$\implies \text{as $x \to 3^-$}, F(x) \to \cfrac{1}{4}(3-1) = \cfrac{2}{4} = \cfrac{1}{2}$

$\text{So, at $x = 3$, it is necessary for $F(x) = \cfrac{1}{2}$, explaining the "$\cfrac{1}{2}$+" term in the next piece}$

$\text{For $3 \le x < 7$,}\ F(x) = \cfrac{1}{2} + \cfrac{1}{8}(x-3)$

$\implies \text{as $x \to 7^-$}, F(x) \to \cfrac{1}{2} + \cfrac{1}{8}(7-3) = \cfrac{1}{2} + \cfrac{4}{8} = 1$

$\text{So, for $x \ge 7$, it is necessary for $F(x) = 1$, as a CDF cannot exceed 1}$

So you take them in cases?

specificagent1 · Sep 21, 2021

CM_Tutor said:
If you think about tax scales questions from junior maths, the function we have here is similar, in that each piece has a constant added (the cumulative probability from earlier pieces. Hence

$\text{For $1 \le x < 3$,}\ F(x) = 0 + \cfrac{1}{4}(x-1)$

$\implies \text{as $x \to 3^-$}, F(x) \to \cfrac{1}{4}(3-1) = \cfrac{2}{4} = \cfrac{1}{2}$

$\text{So, at $x = 3$, it is necessary for $F(x) = \cfrac{1}{2}$, explaining the "$\cfrac{1}{2}$+" term in the next piece}$

$\text{For $3 \le x < 7$,}\ F(x) = \cfrac{1}{2} + \cfrac{1}{8}(x-3)$

$\implies \text{as $x \to 7^-$}, F(x) \to \cfrac{1}{2} + \cfrac{1}{8}(7-3) = \cfrac{1}{2} + \cfrac{4}{8} = 1$

$\text{So, for $x \ge 7$, it is necessary for $F(x) = 1$, as a CDF cannot exceed 1}$

1 mark for all this wtf?

CM_Tutor · Sep 21, 2021

specificagent1 said:
1 mark for all this wtf?

I wasn't offering a solution, I was trying to illustrate another way to look at the problem and understand what is going on.

If you want a solution based on this, we can see that the CDF is zero prior to $x=1$ , increases linearly to 0.5 by $x=3$ , and then increases linearly (at a different rate) reaching 1 by $x=7$ . Since the CDF is the integral of the PDF, the latter must be constant in each region and is:

$\begin{align*} f(x) &= 0\ \text{for $x < 1$} \\ &= \cfrac{1}{4}\ \text{for $1 \le x < 3$} \\ &= \cfrac{1}{8}\ \text{for $3 \le x < 7$} \\ &= 0\ \text{for $x \ge 7$} \end{align*}$

Thus...

=)(= said:
So you take them in cases?

Yes, you can approach this by differentiation within each piece / case, and you can see that

$\cfrac{d}{dx} \left[\cfrac{1}{4}(x - 1)\right] = \cfrac{1}{4} \qquad \qquad \text{and} \qquad \qquad \cfrac{d}{dx} \left[\cfrac{1}{2} + \cfrac{1}{8}(x - 3)\right] = \cfrac{1}{8}$

How to do part C (2 Viewers)

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