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Suppose that $Y_1, Y_2$ are independent and from a distribution where:

$$ f_\theta(y) = \begin{cases} \frac{3y^2}{\theta^3} &\text{for $0 < y \leq \theta$}, &\theta>0\\ 0 &\text{otherwise} \end{cases} $$

Suppose that I want to find the pdf of the maximum of them, ie, I want to find $T = max(Y_1, Y_2)$.

To do this, I have:

\begin{align} F_T(t) &= P(T \leq t)\\ &= P(max(Y_1, Y_2) \leq t) \\ &= P(Y_1 \leq t, Y_2 \leq t) \\ &= P(Y_1 \leq t) P(Y_2 \leq t)\\ &= \frac{t^6}{\theta^6} \mathbb{1}(t \leq \theta) \end{align}

I obtained $P(Y \leq t)$ as: $P(Y \leq t) = \int_{-\infty}^{t}f_\theta(y)dy = \frac{t^3}{\theta^3}$

$$ f_T(t) = \frac{d}{dt}\frac{t^6}{\theta^6}\mathbb{1}(t \leq \theta) = \frac{6t^5}{\theta^6} \mathbb{1}(t \leq \theta) $$

My question here is:

1) When taking the derivative of $\frac{t^6}{\theta^6}\mathbb{1}(t \leq \theta)$, how do I take into account the indicator random variable?

2) For $P(Y \leq t) = \int_{-\infty}^{t}f_\theta(y)dy = \int_{-\infty}^{t}\frac{3y^2}{\theta^3}\mathbb{1}(y \leq \theta)dy = \frac{t^3}{\theta^3}\mathbb{1}(t \leq \theta)$, how do I handle the indicator $\mathbb{1}(y \leq \theta)$ within the integral?

user321627
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  • You should consider the cases $t \leq \theta$ and $t > \theta$ separately. – Mur1lo Dec 03 '16 at 19:43
  • @Mur1lo I see, so you are saying that by consider both cases, it becomes a sum and the $t > \theta$ part drops because it's $0$? – user321627 Dec 03 '16 at 20:05
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    That's it. This means that for $t\geq\theta$ $P(Y\leq t)$ is a constant (= 1) – Mur1lo Dec 03 '16 at 20:09
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    Possible duplicate of [What is the intuition behind an Indicator Function?](http://stats.stackexchange.com/questions/239055/what-is-the-intuition-behind-an-indicator-function) – Xi'an Dec 03 '16 at 21:05

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