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. $\endgroup$
    – Minus One-Twelfth
    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". $\endgroup$
    – Minus One-Twelfth
    Apr 9, 2019 at 9:07
  • $\begingroup$ @MinusOne I did it, but I couldn't continue $\endgroup$
    – Fatemeh Asgarinejad
    Apr 9, 2019 at 9:15

2 Answers 2

Reset to default
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. $\endgroup$
    – Fatemeh Asgarinejad
    Apr 9, 2019 at 9:16
  • $\begingroup$ @Fatemehhh You're welcome. $\endgroup$
    – Clement C.
    Apr 9, 2019 at 9:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .