Can anyone identify this distribution?
I'm given a pdf in a different form and asked to identify the random variable. Also, there is a positive constant $c$ that i have to identify.
PDF: $f(x)= c*\exp(-x^2-x/4)\ \forall x\in\mathbb{R} $
I suspect this is the normal distribution but I am not sure.
I have tried to factor the exponent by adding and subtracting 1/64
$$ \begin{align} c\ \exp(-x^2-x/4)&=c\ \exp(-(x^2+x/4+1/64-1/64))\\ &=c\ \exp(-(x^2+x/4+1/64)+1/64)\\ &=c\ \exp(-(x+1/8)(x+1/8)+1/64)\\ &=c\ \exp(-(x+1/8)(x+1/8))*\exp(1/64)\\ &=c\exp(1/64)\exp\big(-(x+1/8)(x+1/8)\big) \end{align} $$
but I keep getting stuck. Does anyone have any ideas?
distributions normal-distribution mathematical-statistics pdf
Sycorax says Reinstate Monica
Joel SinofskyJoel Sinofsky
$\begingroup$ If the domain is the real line, and the term in the exponent is quadratic with leading coefficient negative, it's always normal. Take out a $-\frac12$ in the exponent, complete the square to find the parameters and then the required normalizing constant will be obvious. $\endgroup$ – Glen_b -Reinstate Monica Sep 18 '15 at 1:04
$$ \begin{align} f(x) = c\ \exp({-x^2-\frac{x}{4}}) &= c\ \exp({-(x+\frac{1}{8})^2+\frac{1}{64}})\\ &= c' \exp(-\frac{y^2}{2})\ \forall -\infty < y < \infty \end{align} $$
where $c'=c\ \exp(\frac{1}{64})$ and $y^2=2(x+\frac{1}{8})^2$
$f(y) = c'\ \exp(-\frac{y^2}{2})\ \forall -\infty < y < \infty$ is analogous(assuming $c$ should be such that $f(x)$ is a valid PDF) to a standard normal distribution
and hence $c' = \frac{1}{\sqrt{2\pi}}$ $\implies$ $c = \frac{e^{-\frac{1}{64}}}{\sqrt{2\pi}}$
rightskewedrightskewed
$\begingroup$ For future reference -- with questions that could well be homework (or worse, on a test/exam), please lean toward hints and guidance rather than present fully worked (or nearly so) solutions. $\endgroup$ – Glen_b -Reinstate Monica Sep 18 '15 at 1:08
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