LessonThis is a lesson restated in new words. Any figure in it is an example from the source, not this desk's record.
If three returns are independent and each is uniform on the interval from 0 to 1, the minimum has cumulative distribution function 1 minus (1 minus z) cubed, for z inside that interval. Differentiate and the density is 3 times (1 minus z) squared. The probability that the minimum is at most one half is 1 minus (1/2) cubed, which is 7/8. The trick is to flip the inequality so that “the smallest is above z” becomes “every return is above z,” which factors under independence.
Why it matters
Interview questions and exams often leave the density rules you practiced and hand you a minimum or a maximum instead. A linear rescaling of one variable has a routine Jacobian. A minimum does not look like a routine anything, because the event “the smallest is below z” does not say which of the three variables did the work. The lecture’s setup uses a toy portfolio of a stock, a bond, and a crypto position, with annual returns that are independent and uniform on zero to one. The minimum is the worst of the three. You want probabilities about that worst return. The uniform assumption makes “below zero” a dull question, because these draws are not negative. A cutoff inside the interval, such as one half, is the nontrivial one. If the returns were normal, a cutoff at zero would matter more. The algebra is the same idea.
The idea in plain words
Let Z be the minimum of the three returns. By definition, the distribution function at a point z is the probability that Z is at most z. Substitute the minimum and you have a probability that is true as soon as any one of the three is low. That “any one” event is awkward to write as a product.
Flip it. The probability that the minimum is at most z equals 1 minus the probability that the minimum is greater than z. If the smallest of three numbers is greater than z, then all three are greater than z. The lecture’s picture is a domino: you cannot have a minimum above z unless every input cleared z. The opposite event does not dominate in the same way. A minimum below z can happen because one input dipped, while the other two stayed high. There is no matching three-way statement.
Now independence applies. The probability that all three exceed z is the product of the three separate probabilities. Identical uniforms make those three probabilities equal. On the interval from 0 to 1, a standard uniform has density 1 and distribution function equal to the point itself. The probability of exceeding z is therefore 1 minus z, for z between 0 and 1. The product is (1 minus z) cubed. Subtract from 1 and you have the distribution function of the minimum.
Differentiate in z. The derivative of 1 is zero. The chain rule on (1 minus z) cubed brings down 3, times (1 minus z) squared, times the derivative of (1 minus z), which supplies a minus sign that cancels the minus already sitting in front. The density on the interval is 3 times (1 minus z) squared. Outside the interval the uniform inputs cannot land, so the minimum’s density is zero there too.
How to use it
Write the distribution function as a probability, substitute the min or the max, and flip before you expand. Only then multiply, and only if the variables are independent. If they are not independent, the product step is false and you still have a joint probability the lecture does not solve.
What usually goes wrong
Expanding “minimum at most z” directly into “each variable at most z” is the error the flip is designed to prevent. That statement is stronger than the event and gives the distribution of the maximum’s cousin, not the minimum. One low draw is enough to pull the minimum down.
Questions people ask
What is the distribution function of the minimum?
For z between 0 and 1, 1 minus (1 minus z) cubed. It is 1 minus the chance that all three independent uniforms exceed z, and that chance is (1 minus z) cubed.
What is the density?
3 times (1 minus z) squared on zero to one, and zero outside. It is the derivative of the distribution function.
What is the probability the minimum is at most one half?
7/8, because 1 minus (1/2) cubed is 7/8. That uses the uniform support in the lecture’s toy portfolio, not a fitted market distribution.