Can anyone identify this distribution?

Question

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?

Answer 1

$$ \begin{align} f(x) = c\ \exp({-x^2-\frac{x}{4}}) &= c\ \exp({-(x+\frac{1}{8})^2+\frac{1}{64}})\\ &= c' \exp(-\frac{y^2}{2})\ \forall -\infty < y < \infty \end{align} $$

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 $c' = \frac{1}{\sqrt{2\pi}}$ $\implies$ $c = \frac{e^{-\frac{1}{64}}}{\sqrt{2\pi}}$