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A die is cast $3$ independent times. Let $X_i$ be the random variable representing the number on the face appearing at the $i$th cast. Let $Y$ be the random variable defined by $Y = \max(X_1,X_2,X_3)$. Find the C.D.F and P.D.F of $Y$.

Answer key gives: $g(y) = \frac1{6^3} (y^3 - (y-1)^3)$, $0<y<1$ as the p.d.f.

What I did was the following:

Assign to each $X_i$ the marginal density function: $f_i(x_i) = \frac16$ for $x_i \in \{1,..., 6\}$ and $0$ otherwise. Then because the $X_i$'s are independent and if $G$ is C.D.F of $Y$:

$$G(y) = P[Y \le y] = P[X_k \le y, k] = P(X_1 \le y) \times P(X_2 \le y) \times P(X_3 \le y) = \left( \sum_{x_k \le y} f_k(x_k) \right)^3 = \left( \frac{y}{6} \right)^3 = \frac1{6^3} y^3, 0 \le y < 1$$

So that $g(y) = \frac3{6^3}y^2$ which differs by $\frac1{6^3}(1 - 3y)$ from the answer key.

What did I do wrong?

George
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1 Answers1

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So far you've calculated the C.D.F.. In order du calculate the probability mass function you now need to use the identity $$P(Y = x) = P(Y \le x) - P(Y < x)$$

Since this random variable is $\mathbb{N}$-valued, we have $P(Y < x) = P(Y \le x - 1)$ which yields the assertion.

Dominik
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  • But isn't the probability mass function just the derivative of the C.D.F? – George Aug 08 '15 at 12:52
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    This is only the case for densities with respect to the lebesgue-measure. Since this random variable is $\mathbb{N}$-valued, you need to calculate the density with respect to the counting measure. – Dominik Aug 08 '15 at 12:55
  • right! what a mistake.. thanks a lot. – George Aug 08 '15 at 12:57
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    @George It is common to use the abbreviation PMF (probability mass function) in a cases like this, and to use abbreviation PDF (probabilitity density function) when you are dealing with a density wrt Lebesgue measure. – drhab Aug 08 '15 at 13:03