Let $X, Y$ be zero mean, unit variance Gaussian random variables with correlation $\rho = -1/2$, find the PDF of $Z = X+Y$
Here is what I have done so far: (Note, multivariate normal distribution)
(Assume the first equality is given)
\begin{align} f_Z(z) &= \int\limits_{-\infty}^{\infty} f_{XY}(x, z - x)dx\\ & = \int\limits_{-\infty}^{\infty} \dfrac{1}{2\pi\sqrt{1-\rho^2}} e^{\dfrac{-(x^2 - 2\rho x(z-x) + (z-x)^2)}{2(1-\rho^2)}} dx\\ & = \int\limits_{-\infty}^{\infty} \dfrac{1}{2\pi\sqrt{1-\rho^2}} e^{\dfrac{-(x^2 + x(z-x) + (z-x)^2)}{2(3/4)}} dx\\ & = \int\limits_{-\infty}^{\infty} \dfrac{1}{2\pi\sqrt{1-\rho^2}} e^{\dfrac{-(x^2 - zx + z^2)}{2(3/4)}} dx\\ \end{align} where we plugged in $\rho = -1/2$ in the integrand
Now a hint instructs me to perform complete the square, I am not sure what this means, because:
$x^2 - zx + z^2 = x^2 - 2zx + z^2 + zx = (x-z)^2 + zx$
Is this completing the square? Because I am getting no where with this integral: \begin{align} f_Z(z) &= \int\limits_{-\infty}^{\infty} f_{XY}(x, z - x)dx\\ & = \int\limits_{-\infty}^{\infty} \dfrac{1}{2\pi\sqrt{1-\rho^2}} e^{\dfrac{-((x-z)^2 + zx)}{2(3/4)}} dx\\ \end{align}
The final expression is one that consist entirely of $z$ only, I do not see how I can get rid of the $x$ in the above integral.
Can someone clarify what I need to do?
Note: following the answer
\begin{align} f_Z(z) &= \int\limits_{-\infty}^{\infty} f_{XY}(x, z - x)dx\\ & = \int\limits_{-\infty}^{\infty} \dfrac{1}{2\pi\sqrt{1-\rho^2}} e^{\dfrac{-((x-\frac{z}{2})^2 + 3z^2/4}{2(3/4)}} dx\\ \end{align}
Factoring out the $z^2$ gives the desired solution
The question is how would one realize to complete $x^2 - zx + z^2$ into $(x-\frac{z}{2})^2 + 3z^2/4$?
Define $y=x-\frac{z}{2}$ so $x^2-zx+z^2=y^2+\frac{3z^2}{4}$.