Jane saw a police officer and Roger saw one too Statement:

Jane saw a police officer and Roger saw one too

Let

*

*$J(x)$: Jane saw police officer $x$


*$R(x)$: Roger saw police officer $x$
Hence,
$$\exists x \exists y \left( J(x) \land R(y) \right)$$
Is this correct?
 A: Your translation is not incorrect, but quite imprecise. What you summarized as one predicate, "$P(x)$ = Jane saw police officer ", still contains an individual (Jane), a relation (saw), and a property (is a police officer). But it is not good practice to import specific individuals (Jane) and additional restrictions (is a police officer) into a predication (z saw y). To exaggerate, you could of course just shove everything into one symbol and define a single zero-place predicate "$P()$ = Jane saw a police officer and Roger saw one too" and be done with it -- technically not wrong, but kinda defeats the point of it all. Atomic formulas should express atomic propositions; what can be decomposed in more detail should be decomposed in more detail.
So analyze the sentene further into the parts mentioned above:

*

*individuals:

*

*$j$ = Jane

*$r$ = Roger



*properties and relations:

*

*$po(x)$ = $x$ is a police officer

*$saw(x,y)$ = $x$ saw $y$
As a rule of thumb,
$$\text{an A (is) B/there is an A that B}$$
translates as
$$\exists x (A(x) \land B(x))$$
and
$$\text{all A (are) B/everything that A B}$$
translates as
$$\forall x (A(x) \to B(x))$$
What you want here is "$\text{ a } \underbrace{\text{police officer}}_{A} \text{ is } \underbrace{\text{seen by Jane}}_{B}$", and the same with Roger. So you end up with
$$\exists x (po(x) \land saw(j,x)) \land \exists y (po(y) \land saw(r,y))$$
If you prefer, you can write the formula in a logically equivalent variant with the quantifiers moved to the front and the conjuncts commuted:
$$\exists x \exists y (po(x) \land po(y) \land saw(j,x) \land saw(r,y))$$
What is important is, as you did, to introduce two different individuals for the police officers and apply the predications to both of them, because the original sentence doesn't say that Jane and Roger necessarily saw the same officer (though the above formula correctly doesn't exclude this possibility, either).
A: If the question was "Jane saw every police officer", would $\lnot \exists x . \lnot P(x)$ be correct answer?  That is saying "there is not a police officer that Jane didn't see" which while technically equivalent, isn't actually the hypothetical question.
$$\exists x \exists y \left( J(x) \land R(y) \right)$$
This is technically correct, but it is as if you were translating "there were some police officers seen by Jane and Roger."  The statement is "Jane saw a police officer and Roger saw one too", so you would expect the answer to be something like $\text{this } \land \text{ that}$.  I would suggest:
$$(\exists x. J(x)) \land (\exists x.R(x))$$
A: I wouldn't call your solution incorrect per se but it's not an ideal answer. The functions should be predicates, which are essentially clear yes/no questions. The desired predicate (named $P$ in the comments) asks "did $x$ see a police officer", to which "yes" implies $x$ did, while "no" implies $x$ didn't. But your predicate, $J$, asks "did Jane see that police officer" - it's not as informative what yes or no mean here. "No" could mean Jane didn't see the police officer (but someone else did) or it could mean Jane didn't see that police officer (but she did see a different one). This is why $P$ is better.
