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Help with Probability. Miller Chapter 2 Practice Question 2.

Alex_1

Active Member
Hi, you have in this question p = F(x) = (3/125)*(1/3)*x^3 = x^3/125 and you need to solve for x. The inverse of the third power (x^3) is the cube root or "^(1/3)".

So from p = x^3/125, you have 125*p = x^3 and from there you apply the cube root to both sides of the equation resulting in:
(125*p)^(1/3) = x^3^(1/3) --> (125*p)^(1/3) = x.

I hope this helps. Thanks.
 

sagrr

New Member
Can someone briefly explain the theory behind solving for x and how thats related to the inverse CDF?
 

Nicole Seaman

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@CYLoh,

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David Harper CFA FRM

David Harper CFA FRM
Staff member
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Hi @sagrr

It's a good question because this connection is the essence of VaR. It may help (it helps me, is what I can say!) to recollect with the normal distribution:
  • We could write CDF of the standard normal as N(.) = N[q(z)] = p; e.g., N(-2.33) = 1% and N(-1.645) = 5%. I am using q(z) to remind that this z-value is a distributional quantile, as in, the quantile we happen to call the z-value in a standard normal distribution. The (p) is significant because it reminds us that a CDF is a probability bounded by 0% and 100%. In this way, N[q(z)] = p is like N(x) = p; i.e., in a CDF each quantile corresponds to a probability, as -2.33 corresponds to 1.0%.
  • In 300.2, we are given the f(x), a pdf, which we integrate in order to retrieve F(x), its implied CDF. Most F(x) will not be CDFs, but this one is designed to be a probability so this F(x) = p. Similar to standard normal N[q(z)] = p , but instead it is a custom, very non-normal, function.
  • So we have a function F(x) = x^3/125, which is just a function ... but it's a special function: it's a CDF (cumulative probability) function, such that F(x) = p, which is the point. Now F(x) = p = x^3/125. The probability is directly linked to the quantile. If this were a normal distribution, we'd use the inverse N[-1](p) = q(z) to retrieve the quantile associated with a given probability; e.g, N[-1](5%) = -1.645.
  • Our desired VaR is 95%, so our p is 5%, and we want F[-1](5%) = q(x); i.e., what quantile (value of x) satisfies the Pr[X < q(x)] = 5%. Solving for (x) as a function of p = 5% returns the 5% quantile: the value of x which should be randomly exceeded 95% of the time. I hope that helps,
 

sagrr

New Member
Can you explain to me why in 300.2, I use .05 for p, while in 300.3, I use .95. Both ask for 95% VaR. Am I missing some nuance in wording or an important concept?
 

Alex_1

Active Member
Hi @sagr, the difference is explained by David in the following post:
https://www.bionicturtle.com/forum/threads/p1-t2-300-probability-functions-miller.6728/#post-24058
Basically, for 300.2 you set p = 0.05 because it is stated "that 5.0% of the distribution is less than or equal to (x), what is this price", the focus being on "less than", which implies a quantile of 5%, hence a VaR of (1-5%) or 95%.
For 300.2 as David states, the key word is "loss severity" which is a function of losses and not of the price which inverts the perspective, hence you require a probability of 95% and VaR of (1-95%) or 5%.

Let me know if this helps.
 

David Harper CFA FRM

David Harper CFA FRM
Staff member
Subscriber
Thanks @Alex_1 great answer. In my opinion, this is not nuance, this is a basic but meaningful idea: we can't try to memorize (get mechanical about) left or right tail. It is either depending on the context and scenario. Sometimes the distribution is prices (where the risk is realized in the left tail), but often the distribution losses (where risk is realized in the right tail). Thanks!
 
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