# Are the square and the maximum of distribution functions a distribution function?

Let $F$ and $G$ be (one dimensional) distribution functions. Decide which of the following are distribution functions.

(a) $F^2$,

(b) $H$, where $H(t) = \max \{F(t),G(t)\}$.

I know the definition and properties of distribution function but I could not solve the problem in rigid way

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Hint: If $X$ and $Y$ are independent random variables with identical cumulative probability distribution function (CDF) $F(t)$, and $Z = \max\{X,Y\}$, what is the CDF of $Z$, that is, what is $F_Z(t) = P\{Z \leq t\}$? – Dilip Sarwate Jul 15 '12 at 13:08
@Dilip: Yes, that is for (a). Now, (b)... :-) – Did Jul 15 '12 at 17:23
@did Perhaps after the OP responds to Zev Chonoles's request for a re-write or showing some work... – Dilip Sarwate Jul 15 '12 at 22:36
@Dilip: Quite a good strategy, which I fully support. – Did Jul 15 '12 at 22:43

A function $F$ is a cumulative probability distribution function on $\mathbb{R}$ if and only if the following are true:

• $F(x)\to0$ as $x\to-\infty$;
• $F(x)\to1$ as $x\to+\infty$;
• $F$ is non-decreasing, i.e. whenever $a<b$ then $F(a)\le F(b)$.
• $F$ is right-continuous.

So ask yourself whether those are true of $F^2$ and of $\max\{F,G\}$ if they are true of $F$ and $G$.

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This is wrong. For example $F(x)=0$ if $x\lt0$, $F(x)=\frac12$ if $0\leqslant x\leqslant 3$ and $F(x)=1$ if $x\gt3$ is not a CDF although the three bulleted properties hold. – Did Jul 15 '12 at 6:40
It should be right continuous as well. Your example is not at $x=3$. – Seyhmus Güngören Jul 15 '12 at 10:04
@Seyhmus: Yes. (But who are you talking to? The second sentence seems to be addressed to me and the first one to Michael.) – Did Jul 15 '12 at 11:32
I wrote the comment to you. Indicating that you gave a nice example to make the properties described by Michael complete. – Seyhmus Güngören Jul 15 '12 at 11:40
@Seyhmus: I see. (And please use the @ thing.) – Did Jul 15 '12 at 16:54

(a)$F^2$: Since $F$ is non-decreasing ,$F^2$ is also non-decreasing. Since $F$ is right continuous,$F^2$ is also right continuous and Since $F(X)\to 0$ as $x\to -\infty$ ,$F^2(X)\to 0$ as $x\to -\infty$ and lastly since $F(X)\to 1$ as $x\to \infty$ $F^2(X)\to 1$ as $x\to \infty$.Hence $F^2$ is a distribution function.

(b)Similarly, $H$ is also satisfying all criteria of distribution functions.Hence $H$ is a distribution function.

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ultimately I understand that the problem.I am stuck because I am not writing anything .Whenever I start writing it is solved . thank you all. – Argha Aug 4 '12 at 4:11