# Shortcut to finding the distribution of a specific random variable

Question: A dice is rolled 3 times. Let X denote the maximum of the three values rolled. What is the distribution of X (that is, P[X = x] for x = 1,2,3,4,6)? You can leave your final answer in terms of "x". [Hint: Try to first compute P[X ≤ x] for x = 1,2,3,4,5,6].

Solution: Let $$X$$ denote the maximum of the three values rolled. We are interested in $$P(X = x)$$, where $$x = 1,2,3,4,5,6$$. First, define $$X_1, X_2, X_3$$ to be the values rolled by the first, second, and third dice. These random variables are independent and identically distributed between 1 and 6 inclusive. Following the hint we first compute $$P[X ≤ x]$$ for $$x = 1,2,3,4,5,6$$:

$$P(X ≤ x) = P(X_1 ≤ x)P(X_2 ≤ x)P(X_3 ≤ x) = \left(\frac x6\right)\left(\frac x6\right)\left(\frac x6\right) = \left(\frac {x}{6}\right)^3$$

Then, observing that $$P(X = x) = P(X ≤ x)−P(X ≤ x−1),$$

$$P(X = x) = \left(\frac {x}{6}\right)^3 - \left(\frac {x-1}{6}\right)^3 = \left(\frac {3x^2-3x+1}{216}\right)^3$$

And using the above formula, we can find $$P(X = x)$$ for all values of $$x$$. My question is how did we come up with the equality in the first equation? Why is the following true?:

$$P(X ≤ x) = P(X_1 ≤ x)P(X_2 ≤ x)P(X_3 ≤ x)$$

First note that the event $$(X\leq x)=(X_1\leq x)\cap (X_2\leq x)\cap (X_3\leq x)$$ which basically is a fact about numbers namely that $$\max_{i=1}^n(x_i)\leq x$$ iff $$x_i\leq x$$ for all $$i$$. Then use independence to write $$P(X\leq x)=P((X_1\leq x)\cap (X_2\leq x)\cap (X_3\leq x))=P(X_1\leq x)P(X_2\leq x)P(X_3\leq x)$$
If $$x_1,x_2,x_3$$ are real numbers then $$\max \{x_1,x_2,x_3\} \leq x$$ iff $$x_i \leq x$$ for each $$i$$. So $$P(X \leq x)=P(X_1 \leq x,X_1 \leq x,X_1 \leq x)=P(X_1 \leq x) P(X_2 \leq x)P(X_3 \leq x)$$ by independence.