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Uniform Random Varibable CFA Quant!!

Page 388 of the CFA Quant book under solution to 1: EBITDA/Interest is a continous uniform random variable because all outcomes are equally likely.
I thought that was the definition for a discrete uniform random variable ( probabilities are equal for all possible outcomes).
I’m confused about Discrete uniform random variables and continuous uniform distributions.
Can someone please help me.

Hi dudeinthecity
Here are some my ideas.
Discrete random variable is variable that its possible outcome can be counted. Continuous random variable is variable that its possible outcome can not be counted.
Discrete uniform random variable is one for which gives the equal possible outcome for each variables.
Discrete distribution gives the finite probability of random variable at the point occur but continuous distribution cannot give the finite probability of continuous variable at exact the point it occur even possible outcome occurs at that point. Continuous distribution has meaning value in a range value of random.
As your above example of Ebitda/Interest it is explained as the possible outcomes are too large that even Ebitda/Interest is discrete distribution it is still treated as continuous distribution.
That’s all my though. Hope somebody else will correct our understanding if anything wrong. Thank you.

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The problem says… is the TIE Ratio falls below 2.0 the company will violate its covenants.
You forecast interest charges of 25 million. you estimate of EBITDA runs from [40  60] million.
1) if the outcomes for ebitda are equally likely, what is the probability that EBITDA/interest will fall below 2.0, breaching the covenant.
So………
I calculated my TIE range 40/25 = 1.6 60/25 =2.4
What is confusing me is 2.4 to 1.6 is this the total space???
and why is the probability of falling below 2.0 defined as [2.0 to 1.6]
the right answer is:
2.0 1.6/ 2.41.6 = .5 i just dont get why there is a 1.6 in the numerator….. is this just a rule of uniform random variables?

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a uniform distribution is akin to finding the area of the rectangle covered by your situation.
In this case  full rectangle = 1.6 2.4
TIE falls below 2 rectange is 1.6 2.0
so your probability = .4 / .8 = .5
All possibilities are equally likely…. that is what it means.

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