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I have a CDF $\ F(x)= \frac {x+1}{k} $ for $\ -1 \leq x \leq 1 $

Asked to find the pdf results in $\ f(x) = \frac {1}{k} $

Then asked to find the value of the constant k leads me to re-integrate:

$\int_{-1}^{1}\frac {1}{k} \text{ d}x $

$\int_{0}^{1}\frac {x}{k} \text{ d}x + \int_{-1}^{0}\frac {x}{k} \text{ d}x $

$\ [ \frac{1}{k} + 0] + [0 + \frac{-1}{k}] = 1$

Which results in k =0. Naturally I am suspicious.

Can anyone verify these steps?

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    $\begingroup$ Your (wrong) result does not imply that $k=0$. Since the integral would be always 1, k could be everything except 0. $\endgroup$ – callculus Mar 16 '17 at 7:17
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You computed the integral incorrectly. $$\int_{-1}^1 \frac{1}{k} \mathop{dx} = \frac{1}{k} \int_{-1}^1 1\mathop{ dx} = \frac{2}{k},$$ so $k=2$.

You could actually get the value of $k$ directly from the CDF, from the property that $F(x) \to 1$ as $x \to \infty$. (In particular, you need $F(1)=1$.)

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