$2$-fiber product is a scheme then map of stacks is representable Ariyan Javanpeykar said here in comments that,

$X\times_{\mathcal{X}}X$ being a scheme is equivalent to representability of $X\rightarrow \mathcal{X}$.

Context is as in this question.
Suppose $p:X\rightarrow \mathcal{X}$ is representable, then, for any scheme $M$ with a map of stacks $M\rightarrow \mathcal{X}$, the $2$-fiber product $X\times_{\mathcal{X}}M$ is a scheme. In particular, we can take $M=X$ then, $X\times_{\mathcal{X}}X$  is a scheme. Thus, $p:X\rightarrow \mathcal{X}$ is representable implies $X\times_{\mathcal{X}}X$ is a scheme.
Suppose it is given that $X\times_{\mathcal{X}}X$ is a scheme. Then, I want to see that $X\rightarrow \mathcal{X}$ is representable i.e., hen, for any scheme $M$ with a map of stacks $M\rightarrow \mathcal{X}$, the $2$-fiber product $X\times_{\mathcal{X}}M$ is a scheme. I wanted to write something like $X\times_{\mathcal{X}}M=(X\times_{\mathcal{X}}X)\times_X M$. But it does not make sense as there is no map $M\rightarrow X$.
any comments on how to see this are welcome.
 A: This is not true even if $\mathcal X$ is an Artin stack. For example, let $G$ be a smooth group scheme over the base $T$, and let $\mathbf BG$ be its classifying stack (the category of $G$-torsors fibered over $Sch/T$). Then $T\times_{\mathbf BG}T=G$ is a scheme. However, the atlas $T\to \mathbf BG$ is not representable by schemes in general. An example where $G$ is an elliptic curve over a normal local scheme $T$ of dimension 2 can be found in the article Ineffective descent of genus one curves by Wouter Zomervrucht (arXiv:1501.04304).
What is true is that if $X$ is an algebraic space, $f:X\to \mathcal X$ is an effective epimorphism of fppf sheaves of groupoids on $Sch/T$, and $X\times_{\mathcal X}X$ is an algebraic space, then $f$ is representable by algebraic spaces. Indeed if $U$ is any algebraic space and $U\to\mathcal X$ any map, then $U\times_{\mathcal X}X\to U$ is representable by algebraic spaces fppf-locally on $U$ (namely after base change to $U\times_{\mathcal X}X$, where it becomes a pullback of $X\times_{\mathcal X}X\to X$), hence it is representable by algebraic spaces because algebraic spaces satisfy fppf descent (which is a somewhat nontrivial fact, see https://stacks.math.columbia.edu/tag/04SJ).
A: Following lemma is from Kai Behrend and Ping Xu's paper (page $8$, lemma $2.11$)  Differentiable Stacks and Gerbes.

Let $f:\mathcal{X}\rightarrow \mathcal{Y}$ be a morphisms of stacks. Suppose given a manifold $U$ and a morphism $U\rightarrow \mathcal{Y}$ which is an epimorphism. If the fiber product $\mathcal{X}\times_{\mathcal{Y}}U$ is representable and $V\rightarrow U$ is a submersion, then $f$ is a representable submersion.

Take $\mathcal{X}=X=U$ in above lemma. Then above lemma turns to 

Let $f:X\rightarrow \mathcal{Y}$ be a morphisms of stacks. Suppose that  $f: X\rightarrow \mathcal{Y}$ is an epimorphism. If the fiber product $X\times_{\mathcal{Y}}X$ is representable and $X\times_{\mathcal{Y}}X\rightarrow X$ is a submersion, then $f$ is a representable submersion.

I guess  relaxing the condition  $X\times_{\mathcal{Y}}X\rightarrow X$ is a submersion conlcudes that $X\rightarrow \mathcal{Y}$ is representable. 
So, 

Consider a  morphism of stacks $f:X\rightarrow \mathcal{Y}$. Suppose that   $X\times_\mathcal{Y}X$ is representable. If further $f:X\rightarrow \mathcal{Y}$ is an epimorphism  then $X\rightarrow \mathcal{Y}$ is a representable submersion. 

This is very much like Marc Hoyois's answer which says that adding the condition that the map is an epimorphism (along with algebraic spaces condition) concludes that the map is representable. 
