0
$\begingroup$

The formula for pdf of normal distribution is

$$ f(x)= \frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(x-\mu)^2}{2\sigma^2}}$$

and I have a pdf that is: $$ke^{-x^2-7x}$$

But I face a paradox in calculating $\mu$

From the formula:

$ {2\sigma^2} = 1 $ ______So_____ ${-x^2-7x} = {-(x-\mu)^2}$ and this gives $\mu = 0$ and then there won't be any -7x

can somebody declare it for me?

$\endgroup$
4
  • $\begingroup$ The main trick is to complete the square in the exponent. $\endgroup$ Apr 9, 2019 at 8:50
  • $\begingroup$ @MinusOne-Twelfth Indeed, but the comment after the two answers stating exactly that loses a bit of impact :) $\endgroup$
    – Clement C.
    Apr 9, 2019 at 9:03
  • $\begingroup$ I mainly put it there in case someone reading wants a "quick highlight". $\endgroup$ Apr 9, 2019 at 9:07
  • $\begingroup$ @MinusOne I did it, but I couldn't continue $\endgroup$ Apr 9, 2019 at 9:15

2 Answers 2

1
$\begingroup$

$ke^{-x^{2}-7x}=ke^{49/4} e^{-(x+\frac 7 2)^{2}}$. So $\mu =-\frac 7 2,\sigma=1/{\sqrt 2}$ and, for the given function to be a pdf, $k$ must be such that $ke^{49/4}=\frac 1 {\sqrt {\pi }}$.

$\endgroup$
1
0
$\begingroup$

Completing the square yields $$ -(x^2+7x) = -\left(x+\frac{7}{2}\right)^2+\frac{49}{4} $$ so $\mu$ can only be $\mu = -7/2$. With that in hand, $$ k e^{-(x^2+7x)} = k e^{-\left(x+\frac{7}{2}\right)^2+\frac{49}{4}} = (ke^{\frac{49}{4}}) e^{-\left(x+\frac{7}{2}\right)^2} = k' e^{-\left(x+\frac{7}{2}\right)^2} $$ where $k' = ke^{\frac{49}{4}}$. Can you finish from there?

$\endgroup$
2
  • $\begingroup$ Thank you so much. $\endgroup$ Apr 9, 2019 at 9:16
  • $\begingroup$ @Fatemehhh You're welcome. $\endgroup$
    – Clement C.
    Apr 9, 2019 at 9:47

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.