# Two Probability Questions from Quantitative Finance Interview Book

I posted the two questions in math stack exchange one month ago but cannot get an answer, so I post it here and appreciate your advice:)

I'm reading an interview book called A Practical Guide to Quantitative Finance Interivew / Chapter 4 Probability Theory. So I ask those following questions (I highlighted my doubts in bold) in this Mathematics Section:

1. Assume that $$X_1, X_2, ...$$ and $$X_n$$ are independent and identically-distributed random variables with uniform distribution between 0 and 1. What is the probability that $$S_n = X_1 + X_2 +....+X_n\leq 1$$?

Solution to question 1: When $$n = 1, P(S_1\leq1)$$ is 1. As shown in Figure 4.6 when $$n=2$$, the probability that $$X_1+X_2\leq1$$ is just the area under $$X1+X2\leq1$$ within the square with side length 1 (a triangle). So $$P(S_2\leq1) = 1/2$$. When $$n=3$$, the probability becomes the tetrahedron ABCD under the plane $$X_1+X_2+X_3\leq1$$ within the cube with side length 1. The volume of tetrahedron ABCD is $$1/6$$ So $$P(S_3\leq1) = 1/6$$ Now we can guess that the solution is $$1/n!$$ To prove it, let's resort to induction. Assume $$P(S_n\leq1) = 1/n!$$. We need to prove that $$P(S_{n+1}\leq1) = 1/(n+1)!$$. Here we can use probability by conditioning. Condition on the value of $$X_{n+1}$$, we have $$P(S_{n+1}\leq1) = \int_0^1f(X_{n+1}))P(S_n\leq1-X_{n+1})dX_{n+1}$$, where $$f(X_{n+1})$$ is the probability density function of $$X_{n+1}$$, so $$f(X_{n+1})=1$$. But how do we calculate $$P(S_n\leq1-X_{n+1})$$? The cases of $$n=2,n=3$$ have provided us with some clue. For $$S_n\leq1-X_{n+1}$$ instead of $$S_n\leq1$$, we essentially need to shrink every dimension of the n-dimensional simplex from 1 to $$1-X_{n+1}$$. So it's volume should be $$(1-X_{n+1})^n/n!$$ instead of $$1/n!$$. So my doubt is: I don't understand why shrinking every dimension of the n-dimensional simplex from 1 to $$1-X_{n+1}$$ gives the result $$(1-X_{n+1})^n/n!$$? What is the reasoning behind this?

1. Let $$X_1$$ and $$X_2$$ be independent and identically distributed random variables with uniform distribution between 0 and 1, $$Y = min(X_1,X_2), Z = max(X_1,X_2)$$. What is the cumulative distribution function of $$YZ$$:

Solution to question 2: when $$0\leq z\leq1, 0\leq y\leq z$$, $$F(y,z)$$ is the shadowed area in Figure 4.7 I don't know why the shadowed area in the screenshot? represented $$F(y,z)$$

• for the second question, you can make your life easier by realising that $\max (X_1, X_2) \cdot \min (X_1, X_2) = X_1 \cdot X_2$, since one of them must be the min, and the other the max, so it's always just their product. – will Jan 5 at 12:08
• @will, thanks a lot! It indeed makes my life much easier:) – M00000001 Jan 6 at 14:53

I think in your book they prove that $$\mathbb{P}(S_n \leq a)=\frac{a^n}{n!}$$ with $$0 \leq a \leq 1$$, and $$a=1$$ is the particular case.

$$n=0$$ is trivial. By induction, we assume that $$\mathbb{P}(S_n \leq y)=\frac{y^n}{n!}$$ $$\forall y \in [0,1]$$

Let $$a \in [0,1]$$, we calculate $$\mathbb{P}(S_{n+1} \leq a)$$. We use the independence between $$S_n$$ and $$X_{n+1}$$ :

$$\mathbb{P}(S_{n+1} \leq a)=\mathbb{P}(S_{n}+X_{n+1} \leq a)=\int_{0}^{1}P(S_n+x \leq a)dx$$

Notice that $$\int_{0}^{1}P(S_n+x \leq a)dx=\int_{0}^{a}P(S_n+x \leq a)dx+\int_{a}^{1}P(S_n+x \leq a)dx$$

$$S_n$$ is almost surely positive, therefore $$\int_{a}^{1}P(S_n+x \leq a)dx=0$$

if $$0 \leq x \leq a$$, we have $$0 \leq a-x \leq 1$$

$$\int_{0}^{a}P(S_n+x \leq a)dx=\int_{0}^{a}P(S_n \leq a-x)dx=\int_{0}^{a}\frac{(a-x)^n}{n!}dx=\frac{a^{n+1}}{(n+1)!}$$

As for the question 2, We know the joint distribution of $$(X_1,X_2)$$, it is given by the density function $$f_{(X_1,X_2)}(x_1,x_2)=1_{x_1 \in ]0,1[}1_{x_2 \in ]0,1[}$$

$$F(y,z)=P(Y \leq y, Z \leq z)=P(min(X_1,X_2) \leq y, max(X_1,X_2) \leq z)=\int_{\{(x_1,x_2)\in ]0,1[^2 :min(x_1,x_2) \leq y, max(x_1,x_2) \leq z \}}{dx_1dx_2}$$

The number $$\int_{\{(x_1,x_2)\in ]0,1[^2 :min(x_1,x_2) \leq y, max(x_1,x_2) \leq z \}}{dx_1dx_2}$$ is the area of $$\{(x_1,x_2)\in ]0,1[^2 :min(x_1,x_2) \leq y, max(x_1,x_2) \leq z \}$$, which is the shadowed area.

• thanks a lot for the generous help! I know typing those mathematical symbols and formula is a lot of work, I really appreciate your clear, detailed and vivid explanation! – M00000001 Jan 6 at 14:54