Probability Problems 03

Last updated on Aug 14, 2023 Interview

In this page, the problems are more like theoretical.

Table of Contents

Law of total probability

In problems where we need to deal with two random variables, it is usually handy to “control” the variables and reduce the dimension. Here the continuous version of law of total probability offers a bridge. Suppose $A$ is the objective event, and it can be partitioned by variable $X$, then

$$ P(A)=\int_{-\infty}^{\infty} P(A \mid X=x) f_{X}(x) d x $$

See the below two examples how they can be applied. Traditionally, $Z=f(X,Y)$ can be solved by convolution. Here, I find using the law here provides a universal working solution and more intuitive.

Z=X+Y

$Z=X+Y$, where $X$ and $Y$ are independent uniform random variables with range $[0,1]$. What is the PDF of $Z$.

Using the total probability, we are deriving the CDF of the variable $Z$. The key part is to partition the value ranges of $Z$ first, and write separate integration by $X$ values.

Z=XY

$Z=XY$, where $X$ and $Y$ are independent uniform random variables with range $[0,1]$. What is the PDF of $Z$.

If you could not view the embeded pdf file here, it is likely due to the network block in company. In that case, please refer to the attachment here.

A convolution solution

$$ \begin{aligned} f_{Z}(z) &=\int_{-\infty}^{+\infty}\left|\frac{1}{x}\right| f_{(X, Y)}\left(x, \frac{z}{x}\right) d x \\ &=\int_{-\infty}^{+\infty}\left|\frac{1}{x}\right| f_{X}(x) f_{Y}\left(\frac{z}{x}\right) d x \\ &=\int_{0}^{1} \frac{1}{x}(1) f_{Y}\left(\frac{z}{x}\right) d x \\ &=\int_{z}^{1} \frac{1}{x}(1)(1) d x \\ &=[\ln x]_{x=z}^{x=1} \\ &=-\ln z \end{aligned} $$

Summation of combinatorial squared

Prove that $\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)^{2}=\left(\begin{array}{c} 2 n \\ n \end{array}\right)$

This is the same as the following. We are choosing a sum of $n$ items form $2n$ population, no matter how they distributed between.

$$ \sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)\left(\begin{array}{c} n \\ n-k \end{array}\right)=\left(\begin{array}{c} 2 n \\ n \end{array}\right) $$

Mathematical Constant

Why $e$ is important?

The following expression converges to $\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n=e$. This is important in Finance, because of the continuous compounding!
This is also an important series convergence, $e=\sum_{n=0}^{\infty} \frac{1}{n !}=1+\frac{1}{1}+\frac{1}{1 \cdot 2}+\frac{1}{1 \cdot 2 \cdot 3}+\cdots$. You have seen it quite many times in a number of probability problems.
Also, in an integral $\int_1^e \frac{1}{t} d t=1$.

Probability