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!
 A: 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 $Y|X \sim N\left( {0,\frac{1}{{\sqrt {2(1 + {x^2})} }}} \right)$.
Hence, this example shows that the normality of the conditional pdf(s) does not imply that the bi-variate density is also normal.
