Proof.
We omit the proof that (1) implies (2), and that (2) implies (3).
Assume (3). Choose a surjective smooth morphism $V \to \mathcal{Y}$. We are going to show that $V \times _\mathcal {Y} \mathcal{X} \to V$ is a universally closed morphism of algebraic stacks. Let $\mathcal{Z} \to V$ be a morphism from an algebraic stack to $V$. Let $W \to \mathcal{Z}$ be a surjective smooth morphism where $W = \coprod W_ i$ is a disjoint union of affine schemes. Then we have the following commutative diagram
\[ \xymatrix{ \coprod _ i |W_ i \times _\mathcal {Y} \mathcal{X}| \ar@{=}[r] \ar[d] & |W \times _\mathcal {Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \ar[d] \ar@{=}[r] & |\mathcal{Z} \times _ V (V \times _\mathcal {Y} \mathcal{X})| \ar[ld] \\ \coprod |W_ i| \ar@{=}[r] & |W| \ar[r] & |\mathcal{Z}| } \]
We have to show the south-east arrow is closed. The middle horizontal arrows are surjective and open (Properties of Stacks, Lemma 100.4.7). By assumption (3), and the fact that $W_ i$ is affine we see that the left vertical arrows are closed. Hence it follows that the right vertical arrow is closed.
Assume (4). We will show that $f$ is universally closed. Let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks. Consider the diagram
\[ \xymatrix{ |(V \times _\mathcal {Y} \mathcal{Z}) \times _ V (V \times _\mathcal {Y} \mathcal{X})| \ar@{=}[r] \ar[rd] & |V \times _\mathcal {Y} \mathcal{X}| \ar[r] \ar[d] & |Z \times _\mathcal {Y} \mathcal{X}| \ar[d] \\ & |V \times _\mathcal {Y} \mathcal{Z}| \ar[r] & |\mathcal{Z}| } \]
The south-west arrow is closed by assumption. The horizontal arrows are surjective and open because the corresponding morphisms of algebraic stacks are surjective and smooth (see reference above). It follows that the right vertical arrow is closed.
$\square$
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