Ergodicity of joint process
Joint behavior cannot always be deduced from individual behavior. For example, if $X$ and $Y$ are (nondegenerate) random variables with 
finite means, then $P\{X&lt;E[X]\}$ and $P\{Y&lt;E[Y]\}$ both are nonzero. This
is because there must be probability mass to the left of the mean (and to the right of the mean too) except in the degenerate case when $X$ has value $E[X]$ with probability $1$. However, $P\{X&lt;E[X],Y&lt;E[Y]\}$ can be zero. Don't believe me? Consider $X\sim$ Bernoulli$(\frac 12)$ and
$Y=1−X$ which is also Bernoulli$(\frac 12)$. Both have means $\frac 12$
but $P\{X &lt; \frac 12, Y &lt; \frac 12\} = 0$.

Turning to ergodicity, one characteristic of an ergodic process is that
over long periods of time (say duration $T$), a typical sample path has value $\alpha$ or less for approximately $F_X(\alpha)\cdot T$ time units where $F_X(\cdot)$ is the first-order CDF of all the random variables
comprising the process. That is, we can get good approximations to the
value of the first-order CDF at $\alpha$
by observing what fraction of time a sample path spends
below level $\alpha$. So, going back to your question, if $\{X(t)\}$
and $\{Y(t)\}$ are ergodic processes, we cannot deduce that the
joint process is also ergodic, for reasons similar to the example
I gave in the first paragraph above. Now, if you impose conditions
such as $\{X(t)\}$ and $\{Y(t)\}$ are independent processes
in addition to being ergodic processes, then Yes, the joint
process is also ergodic.