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If $F(x)$ is a distribution function of a random variable .

If $G_1(x)=(F(x))^3$ and

$G_2(x)=1-(1-F(x))^5$

which of them is a $d.f$ . My answer is that both of them are $d.f$ because they both are $1$ at $\infty$ and $0$ at $-\infty$ They are right continuous as well And to use the non negativity of$ f(x) $ we see that both are non decreasing.Is it correct?

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    Do they both satisfy the requirements for a (cumulative) distribution function? You should make sure of this. – 2017-02-06
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    i have added this in the question now – 2017-02-06
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    So you are claiming that any function which is $1$ at $\infty$ and $0$ at $-\infty$ is a distribution function? That's not correct... there are other properties which you have to check. – 2017-02-06
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    right continuous as well – 2017-02-06
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    also should be non decreasing – 2017-02-06
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    Cant somebody atleast hint something correct if i am wrong any counterexample – 2017-02-06
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    In general yes you are checking the those conditions. Here is another check for this particular problem: Let $X_i$ be i.i.d. random variables with the common CDF $F$. What is the CDF of $\max\{X_1, X_2, X_3\}$ and $\min\{X_1, X_2, X_3, X_4, X_5\}$? – 2017-02-07
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    Yes the first case gives me $G_1$ and second gives me $G_2$. it uses independence – 2017-02-07

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