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

(a) $F^2$,

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

Justify your answer.

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

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Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many would consider your post rude because it is a command ("Justify"), not a request for help, so please consider rewriting it. –  Zev Chonoles Jul 15 '12 at 1:08
    
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
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2 Answers

up vote 1 down vote accepted

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
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(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
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