# Given the bivariate standard normal density, how to show that $X$ and $Y$ are standard normal densities? [closed]

Given the standard bivariate normal density with correlation coefficient $$\rho$$ for $$X$$ and $$Y$$:

$$f_{X,Y}(x,y)=\frac{1}{2\pi\sqrt{1-\rho^2}}e^{-(x^2-2\rho xy+y^2)/2(1-\rho^2)}$$

Is there a way to work backwards and show that $$X$$ and $$Y$$ are standard normal density without the assumption of independence?

## closed as off-topic by StubbornAtom, José Carlos Santos, Parcly Taxel, Javi, CesareoMay 26 at 10:57

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Integrate out one of the variables.

$$f_X(x) = \int \frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left(-\frac{x^2-2\rho xy+y^2}{2(1-\rho^2)}\right)\,dy$$ $$= \frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left(-\frac{x^2(1-\rho^2)}{2(1-\rho^2)}\right)\int \exp\left(-\frac{\rho^2x^2-2\rho xy+y^2}{2(1-\rho^2)}\right)\,dy$$ Notice that I've pulled out a piece of the exponential that doesn't depend on $$y$$ in order to get the numerator inside the integral to have $$\rho^2x^2$$ in it. $$= \frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left(-x^2/2\right)\int \exp\left(-\frac{(y-\rho x)^2}{2(1-\rho^2)}\right)\,dy$$ Letting $$u=\frac{y-\rho x}{\sqrt{1-\rho^2}}$$, we have $$dx = \sqrt{1-\rho^2}\,du$$, and $$= \frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left(-x^2/2\right)\int \exp\left(-u^2/2\right)\sqrt{1-\rho^2}\,du$$ $$= \frac{1}{2\pi }\exp\left(-x^2/2\right)\sqrt{2\pi} = \frac{e^{-x^2/2}}{\sqrt{2\pi}}.$$

Thus $$X\sim \mathcal{N}(0,1)$$, as desired.

• What was wrong in this answer that made you delete it? I was thinking of doing the same thing. – Alex M. May 26 at 16:02
• @AlexM. There are two main problems with that answer. The first is that $X^*$ there isn't the pushforward of a vector field under a single map. Rather it is at every point the pushforward of a single vector under a different map. This causes the pushforward not to commute with the Lie bracket, and indeed, we should have $[X^*,Y^*]=-[X,Y]^*$ as claimed. The other problem is the counterexample I gave, since $X$ is typically identified with a left invariant vector field, but $X^*$ is right invariant (it's the pushforward by right multiplication), so we don't have $X^*=X$ as I claimed. – jgon May 26 at 16:27

Do you know how to find the density of $$X$$ from the joint density of $$X$$ and $$Y$$ in general (that is, without the requirement that the random variables have the specific density shown above)?? If so, apply the same technique. A couple of tricks that you will need to use are

• the method of completing the square; an integrand will have a quadratic polynomial as the argument of the exponential function, and you will need to massage it to write it as a perfect square plus some constants

• recognizing a function as a (univariate) normal density and setting its integral over the real line to have value $$1$$ without necessarily calculating the integral formally to get $$1$$ as the value of the integral.

And if you don't want to waste your time with all this, just look at the other answers where the calculations are written out explicitly in gory detail.

Yes, just find the marginal pdf by integrating out $$y$$ (or $$x$$). \begin{align*} f_X(x)&=\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm{d}y\\ &=\int_{-\infty}^\infty\frac{1}{2\pi\sqrt{1-\rho^2}}\exp\left[-\frac{x^2-2\rho xy+y^2}{2(1-\rho^2)}\right]\,\mathrm{d}y\\ &=\int_{-\infty}^\infty\frac{1}{2\pi\sqrt{1-\rho^2}}\exp\left[-\frac{(1-\rho^2)x^2+(y-\rho x)^2}{2(1-\rho^2)}\right]\,\mathrm{d}y\\ &=\frac{\exp(-x^2/2)}{\sqrt{2\pi}}\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi(1-\rho^2)}}\exp\left[-\frac{(y-\rho x)^2}{2(1-\rho^2)}\right]\,\mathrm{d}y\\ &=\frac{\exp(-x^2/2)}{\sqrt{2\pi}}\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi(1-\rho^2)}}\exp\left[-\frac{(y-\rho x)^2}{2(1-\rho^2)}\right]\,\mathrm{d}(y-\rho x)\\ &=\frac{\exp(-x^2/2)}{\sqrt{2\pi}} \end{align*}