# Normality of conditional pdf(s) does not imply BVN

Show by means of example that the normality of conditional pdf(s) does not imply that the bivariate density is normal.

I know of the following example that if $\ U,V,W$ and $\ T$ are independently distributed with
$U \sim N(0,1)$
$V \sim N(0,1)$
$W \sim N(0,1)$ and
$T \sim U(0,1)$ ,
then $X = U\sqrt{T} + V\sqrt{1-T} \sim N(0,1)$ and $Y = W\sqrt{T} + V\sqrt{1-T} \sim N(0,1)$ but $(X,Y)$ does not follow $BVN$ distribution.

Can this example be modified to answer the above question? If not, can someone please provide a suitable example.

Thanks!

• @Glen_b Thanks for pointing out. Just corrected my mistake. Commented Dec 1, 2015 at 10:41
• I'm deleting my answer; it doesn't serve as an answer in its current state. Commented Dec 1, 2015 at 13:58
• Something quite similar to your example does work (normal conditionals with non-normal joint) in the trivariate case; there's one in Stoyanov, Counterexamples in Probability, 3rd Ed, p97-98; immediately preceding that example is an example of a non-normal bivariate distribution with normal conditionals. Commented Dec 1, 2015 at 14:14
• @Glen_b I had posted a comment but now it has vanished. I don't have access to the above mentioned book so I improvised my solution a bit to adhere to the requirements. $X| W \equiv X$ as $X$ and $W$ are independent of each other. Similarly, $Y| U \equiv Y$. Can you please suggest if my logic is correct or not? Commented Dec 1, 2015 at 15:26
• I responded to it (to say "yes, X is independent of W and Y is independent of U"), but being a comment under my answer it was deleted when I later deleted my answer for not actually answering the question you asked. Commented Dec 1, 2015 at 15:43

Let us define a bivariate random variable $(X,Y)$ having pdf $f(x,y)$ as: $$f(x,y) = \lambda {e^{ - (1 + {x^2})(1 + {y^2})}}$$ where $- \infty < x < \infty$ and $- \infty < y < \infty$. Clearly, $(X,Y) \nsim BVN$
Now the Marginal distribution of $X$ is obtained as follows: $${f_X}(x) = \int\limits_{ - \infty }^\infty {f(x,y)dy}$$ $$= \lambda {e^{ - (1 + {x^2})}}\int\limits_{ - \infty }^\infty {{e^{ - (1 + {x^2}){y^2}}}dy}$$ Now using the fact that: $\int\limits_{ - \infty }^\infty {{e^{ - a{x^2}}}dx = \sqrt {\frac{\pi }{a}} }$, we have: $${f_X}(x) = \lambda {e^{ - (1 + {x^2})}}\sqrt {\frac{\pi }{{1 + {x^2}}}}$$ Similarly, we can find the Marginal distribution of $Y$. $${f_Y}(y) = \lambda {e^{ - (1 + {y^2})}}\sqrt {\frac{\pi }{{1 + {y^2}}}}$$ Now, $$f(x|y) = \frac{{f(x,y)}}{{f(y)}}$$ $$= \frac{{\lambda {e^{ - (1 + {x^2})(1 + {y^2})}}}}{{\lambda {e^{ - (1 + {y^2})}}\sqrt {\frac{\pi }{{1 + {y^2}}}} }}$$ After simplifying a bit we will finally have, $$= \frac{1}{{\sqrt {2\pi } \left( {\frac{1}{{\sqrt {2(1 + {y^2})} }}} \right)}}{e^{ - \frac{1}{2}{{\left( {\frac{x}{{\frac{1}{{\sqrt {2(1 + {y^2})} }}}}} \right)}^2}}}$$ $$\Rightarrow X|Y \sim N\left( {0,\frac{1}{{\sqrt {2(1 + {y^2})} }}} \right)$$ Similarly, we can show that $Y|X \sim N\left( {0,\frac{1}{{\sqrt {2(1 + {x^2})} }}} \right)$.