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?

  • 1
    $\begingroup$ 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

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 $F(1)=1$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.