Let $Z_1,Z_2,Z_3$ be independent standard normal variables. Find their joint pdf and $P(Z_3>Z_1+Z_2)$.

Question

Let $Z_1,Z_2,Z_3$ be independent standard normal variables. Find their joint pdf and $P(Z_3>Z_1+Z_2)$.

If $X\sim N(\mu,\sigma^2)$, then $M_X(t)=e^{\mu t+\sigma^2t^2/2}$, $\infty<t<\infty$.

Now since each $Z_i$ for $i \in \{1,2,3\}\sim N(0,1) \Rightarrow M_{Z_i}(t)=e^{t^2/2}$, where $-\infty<t<\infty$ and $f(z_i)=\frac{1}{\sqrt{2\pi}}e^{-z_i^2/2}$.

$P(Z_3>Z_1+Z_2)=P(Z_1+Z_2-Z_3<0)$

Let $Y=Z_1+Z_2-Z_3$. Then $M_Y(t)=E(e^{tY})=E[e^{t(Z_1+Z_2-Z_3)}]=E(e^{tZ_1}e^{tZ_2}e^{-tZ_3})$

From independence, we have

$M_Y(t)=E(e^{tY})=E(e^{tZ_1})E(e^{tZ_2})E(e^{-tZ_3})=M_{Z_1}(t)M_{Z_2}(t)M_{Z_3}(-t) \tag1$

But since we're squaring $t$, $M_{Z_3}(-t)=e^{t^2/2}$, so the result of (1) is $(e^{t^2/2})(e^{t^2/2})(e^{t^2/2})=e^{(t^2+t^2+t^2)/2}=e^{3t^2/2}$.

As $M_Y(t)=e^{3t^2/2}$, then $Y \sim N(0,3)$. And because $Y \sim N(0,3)$, it follows that $Y$ is symmetric about $0$. Thus $P(Z_3>Z_1+Z_2)=P(Z_1+Z_2-Z_3<0)=0.5$.

Also from independence, we have $f(z_1,z_2,z_3)=\frac{1}{(2\pi)^{\frac{3}{2}}}e^-\frac{(z_1^2+z_2^2+z_3^3)}{2}$

Thus $P(Z_3>Z_1+Z_2)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{z_1+z_2}^{\infty}\frac{1}{(2\pi)^{\frac{3}{2}}}e^-\frac{(z_1^2+z_2^2+z_3^3)}{2}dz_3dz_2dz_1$

Is this correct?

Answer 1

Wtihout a lot of calculations, the joint pdf is the product of the three marginals and

$$\mathbb{P}[Z_3-Z_2-Z_1>0]=0.5$$

being

$$(Z_3-Z_2-Z_1)\sim N(0;3)$$

Answer 2

Another way to render a probability of $1/2$:

For each ordered triple $(u,v,w)$ there is another with equal density, $(-u,-v,-w)$ such that barring the infinitesimal probability of equality, exactly one of $w>u+v,-w<(-u)+(-v)$ is true. This applies to any independent, distributions symmetric about zero.