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I'm working on a quant interview question from the book called Quant Job Interview Questions And Answers (by Mark Joshi and other authors). I cannot understand the following question(not the answer, but the question itself):

Question 5.2: Suppose an asset takes values from a discrete set $v(j)$ and the probability of $v(j)$ is $p(j)$. Write an algorithm that produces the random variable for this asset from a uniformly distributed random variable.

What is the meaning of "produces the random variable for this asset from a uniformly distributed random variable", can any expert give an example to show what it means? Really appreciate your help!

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The question requires you to provide a method which uses uniform random variables and transforms them to generate realizations of the described asset values.

To give a bit more general answer: this is solved by the inverse transform sampling method. The main idea is to obtain realizations of a random variable $x$ with any given distribution function $F(x)$, by using random numbers $u$ ~ $U(0,1)$ and transforming them.

To do this, you need first need to obtain the distribution function $F(x)$ (in your case you have a probability mass function to begin with), and calculate the inverse cdf $F^{-1}(x)$. Finally, for $u$ ~ $U(0,1)$, the values $x = F^{-1}(u)$ have a distribution $F(x)$.

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  • $\begingroup$ Thanks a lot for the quick response, detailed explanation! Really appreciate it! $\endgroup$ – M00000001 Nov 13 at 17:01
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Say your asset can take the discrete values {1,2,3,4} with probabilities {0.4, 0.1, 0.2, 0.3}.

The question is to derive a sampling procedure that returns either {1,2,3,4} with the right probabilities according to the underlying distribution.

The solution is to use a random uniform variable ($u \sim U(0,1)$)and allocate it based on the following:

if $u < 0.4 \implies 1$
if $u \geq 0.4, u < 0.5 \implies 2$
if $u \geq 0.5, u < 0.7 \implies 3$
if $u > 0.7\implies 4$

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  • $\begingroup$ Thanks a lot for the quick response and great example! Really appreciate it! $\endgroup$ – M00000001 Nov 13 at 17:01

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