# can I maximize sum of two log likelihood?

Suppose I have data $$(y_i,x_i)_{i=1}^n$$ and I know two density functions $$f(y_i;\theta)$$ and $$g(x_i;\theta)$$ -- two density function shares the same parameter $$\theta$$.

I want to use both information to estimate $$\theta$$. I consider log likelihood function $$L(\theta;data) = \sum_{i}\log(f(y_i;\theta)) + \log(g(x_i;\theta)).$$

if I maximize this with respect to $$\theta$$, would this give me a consistent estimator for $$\theta$$?

Sure, you can do this. It depends on how you think $$y$$ and $$x$$ are related. I'll give you an example in which this makes mathematical sense, but is contrived and unrealistic.
If $$x$$ and $$y$$ are independent but share a parameter. For instance, suppose $$x_i \sim N(1, \theta)$$ and $$y_i \sim N(-1, \theta)$$. Let $$x = (x_1,x_2, \ldots, x_n)$$ be an iid sample of size from $$N(1, \theta)$$ and $$y = (y_1,y_2, \ldots, y_n)$$ be an iid sample of size from $$N(-1, \theta)$$. Let $$\mathcal{L}_x(\theta \mid x)$$ be the likelihood function for $$x$$ and $$\ell(\theta \mid x)$$ be the log-likelihood function for $$x$$. We adopt similar notation for $$y$$.
\begin{align} \mathcal{L}(\theta \mid x, y)&= \left\{ \prod_{i = 1}^n \mathcal{L}_x(\theta \mid x_i) \right\} \times \left\{ \prod_{i=1}^n \mathcal{L}_y(\theta \mid y_i)\right\}\\ &=\prod_{i=1}^n \mathcal{L}_x(\theta \mid x_i) \mathcal{L}_y(\theta \mid y_i). \end{align} Using standard properties of logarithms gives us
\begin{align} \ell(\theta \mid x, y) &= \log \mathcal{L}(\theta \mid x, y)\\ &= \log \left\{ \prod_{i=1}^n \mathcal{L}_x(\theta \mid x_i) \mathcal{L}_y(\theta \mid y_i) \right\}\\ &= \sum_{i=1}^n \log \left\{\mathcal{L}_x(\theta \mid x_i) \mathcal{L}_y(\theta \mid y_i) \right\} \\ &= \sum_{i=1}^n \left\{ \log \left[\mathcal{L}_x(\theta \mid x_i)\right] + \left[ \mathcal{L}_y(\theta \mid y_i) \right] \right\}\\ &=\sum_{i=1}^n \left\{ \ell(\theta \mid x_i) + \ell(\theta \mid y_i) \right\} \end{align} In this case I imagine the estimator would be consistent under the usual regularity conditions. That's because we can break the sum up into two sums.