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The question I'm having trouble with is:

A continuous random variable has the CDF:

$\hspace{10mm} F(x) = \left\{\begin{array}{rcl} 0 & & x < 0 \\ cx^2 & &0 \leq x \leq 2 \\ 1 & & x > 2 \end{array}\right\}$

Find the value of the positive constant $c$.

I understand that to get $c$ from a pdf, a simple integral is all it takes. However, it is escaping me on how to get $c$ from a CDF. Any help would be appreciated.

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The CDF must be continuous.

The only possible problem is at $x=2$, to make it continuous there you need $4c=1$ and hence $c=\frac14$.

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  • $\begingroup$ Wow, can't believe it was that simple.... thanks! $\endgroup$ – tdashrom Nov 13 '15 at 1:13
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The essential point is that it says "continuous". There is only one value of $c$ that makes this function continuous at the point where $x=2$.

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