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Please excuse my use of images enter image description here

The answer to this is that no the first function cannot be a pdf, but the second one can with a restriction. All explanations I try to look just confuse me further.

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Why for the first function since we have negative values it cannot be a pdf, but for the second one we are allowed to put a restriction? I calculated the $C$ for both functions; for the first one I got a negative $C$ and for the second one I got a positive $C$, does this have anything to do with the decision?

Also an explanation of what $f(5/2)<0$ means and why we are using it for the second function would be appreciated

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Allan
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  • Both functions cannot be a density. Basically if there is any sign change within the support it cannot be a density. – Karn Watcharasupat Nov 15 '17 at 04:32
  • @KarnWatcharasupat Both my professor and the solution I posted say that we can. – Allan Nov 15 '17 at 04:34
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    You are reading the solution incorrectly. The final sentence, albeit confusing and misleading, is not saying that the second function can be a density. What it is trying to say is this: if $f(x)$ has exactly two roots, one at $x=0$ and another greater than $\frac{5}{2}$, then the function $$g(x)=\begin{cases}Cf(x) & x\in(0,\frac{5}{2})\0&\text{otherwise}\end{cases}$$ can be a density, by choosing $C$ appropriately. For in that case, $f$ does not change sign on the interval $(0,\frac{5}{2})$. But the two functions you're given, $f(x)=2x-x^3$ and $f(x)=2x-x^2$, don't have this property. – symplectomorphic Nov 15 '17 at 04:37
  • @symplectomorphic Is there a corresponding $g(x)$ for the first function as well? – Allan Nov 15 '17 at 04:39
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    What? I used different notation from your solution; I called $g(x)$ what your solution calls $f(x)$. But what you call it is immaterial. Just read carefully what I said. – symplectomorphic Nov 15 '17 at 04:40
  • @symplectomorphic Yes I understand, I'm asking if we can construct a $g(x)$ that could be a density for the first function in my question. In the solution it says simply no for the first function, but for the second function we are able to construct the $g(x)$ that you wrote. If not, what is the reason for that? – Allan Nov 15 '17 at 04:44
  • The whole point is that if $f(x)$ is continuous and has no roots between $0$ and $\frac{5}{2}$, then $f(x)$ can't change sign on that interval, so $$g(x)=\begin{cases}Cf(x) & x\in(0,\frac{5}{2})\0 & \text{otherwise}\end{cases}$$ can be a density. I don't know why the final sentence was tacked on after the second example (perhaps it was meant to be the moral of both examples); it applies just as well to the first example, yes. – symplectomorphic Nov 15 '17 at 04:48
  • @symplectomorphic If it applies just as well to the first example, then why did two independent sources say no for the first function, but gave the restriction for the second? – Allan Nov 15 '17 at 04:52
  • Your sources are stupid and poorly written, that's why. Moreover, it's not a "restriction" for the second function. It's a conclusion or generalization based on the two exercises in total. The point is a general one. It has nothing to do with the second example alone. I've said that many times now. – symplectomorphic Nov 15 '17 at 04:53
  • I got it now, I misread my professor. He implied your solution in a way that made it seem that he was saying no for the first, yes for the second. However, the solution included in my question didn't help me understand this – Allan Nov 15 '17 at 04:55

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