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 4.6image 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 4.7image2 I don't know why the shadowed area in the screenshot? represented $F(y,z)$

  • 1
    $\begingroup$ 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. $\endgroup$
    – will
    Jan 5 '20 at 12:08
  • $\begingroup$ @will, thanks a lot! It indeed makes my life much easier:) $\endgroup$
    – M00000001
    Jan 6 '20 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.

  • $\begingroup$ 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! $\endgroup$
    – M00000001
    Jan 6 '20 at 14:54

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