How come in this problem they don't just raise the initially given pdf to the power of 3?

Why do they first convert it to a cdf, raise it to the power of 3, and then convert it back to a pdf

enter image description herehttps://i.stack.imgur.com/4EBeN.jpg


The reason they don't raise the pdf to the third power to find the joint pdf of 3 variables is a) because that is not how you calculate the joint pdf and 2) they are not looking for the joint pdf of the three variables.

To elaborate on the second point, because that is actually more directly related to the subject of the question:

The question is not interested in the pdf of the three variables $T_{1, 2,3}$ together, but in the minimum $T = \min(T_1,T_2,T_3)$. The CDF is used for this calculation because the CDF describes nicely what we want to know.

To know what the probability of $x$ is to be the minimum of $T_1, T_2,T_3$, we need to know what the probability is that all of $T_1,T_2,T_3$ to be higher than $x$.

The function that describes $\mathbb{P}(T_1 > x) = 1 - \mathbb{P}(T_1 \leq x) = 1-CDF_{T_1}(x)$. So the reason they use the CDF is because it gives the probability that $T_1$ is lower than $x$.

This value is then raised to the power 3 to find the probability for all three $T_1,T_2,T_3$ and is then differentiated back to the pdf, as shown in the picture.

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