Can $A$ and $B$ be independent of each other?

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If event $A$ is dependent on event $R$ and so is even $B$, but there is no direct relation between $A$ and $B$. Can $A$ and $B$ be independent of each other?

To give an example, suppose there is a $\dfrac23$ chance that it rains today, and that the credibility of person $A$ is $\dfrac45$ and the credibility of person $B$ is $\dfrac13$. This implies that the response of person $A$ is dependent on whether it rains and so is the response of person $B$, but can I consider that the response of person $A$ is not at all dependent on the response of person $B$?

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The answer to the general question, of course, is yes.

Regarding the specific example, let $R$ be the event that it is raining, let $T_A$ (respectively $T_B$) be the event that $A$ (resp. $B$) tells the truth, and let $R_A$ (respectively $R_B$) be the event that $A$ (resp. $B$) says that it is raining.

Note that $R_A = (R \cap T_A) \cup (R^\complement \cap T_A^\complement).$

We are given that $P(R) = \frac23,$ $P(T_A) = \frac45,$ and $P(T_B) = \frac13.$

Where we go from here depends on other assumptions.

Interpretation I

Let's suppose that $T_A,$ $T_B,$ and $R$ are all pairwise independent. That is, $A$ is equally likely to lie regardless of the weather, the same can be said for $B,$ and $A$ is equally likely to lie regardless of whether $B$ lies. We can then calculate that $P(R_A) = \frac45\cdot\frac23 + \frac15\cdot\frac13 = \frac35,$ $P(R_B) = \frac13\cdot\frac23 + \frac23\cdot\frac13 = \frac49,$ and $$P(R_A, R_B) = \frac45\cdot\frac13\cdot\frac23 + \frac15\cdot\frac23\cdot\frac13 = \frac29 \neq P(R_A)P(R_B).$$ That is, for the particular events defined with these particularly stringent definitions, the responses of $A$ and $B$ are not independent.

Interpretation II

Suppose we assume that $T_A$ and $T_B$ are independent of each other but we do not assume either is independent of $R.$ Then one possible way to model the problem is that $B$ always says it is not raining. Then $P(T_B) = \frac13,$ as required, but $R_B$ is independent of any other event, including $R_A.$

Under this interpretation, the following statement is false:

This implies that the response of person $A$ is dependent on whether it rains and so is the response of person $B.$

Interpretation III

Suppose we assume that $T_A$ and $T_B$ are independent of each other, we do not assume either is independent of $R,$ but we forbid any of the events individually named above from having probability $0$ or $1.$ One possible model then assumes $T_A$ is independent of $R$ and sets $$P(R, R_A, R_B) = P(R, R_A^\complement, R_B) = \frac1{60}$$ and $$P(R^\complement, R_A, R_B) = P(R^\complement, R_A^\complement, R_B) = \frac1{30}.$$ The probabilities of other events are then determined, it is still true that $P(T_B) = \frac13,$ and $R_B$ is independent of $R_A.$

This is easiest to model if $B$ waits to hear $A$'s response and then randomly chooses a response with a probability depending on whether it is raining and whether $A$ said it is raining. Both the responses of $A$ and $B$ depend on whether it is raining, and whether each tells the truth is dependent in probability on whether the other does, but the things $A$ and $B$ actually say ("raining" or "not raining") are independent of each other.

Conclusion

The answer to this question,

If event $A$ is dependent on event $R$ and so is even $B$, but there is no direct relation between $A$ and $B$. Can $A$ and $B$ be independent of each other?

is yes in general. It can be yes for the example given in the question. But we can make some plausible additional assumptions in that example that cause the answer to be no under those additional constraints.

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Throw two fair coins.

Let $A$ be the event that coin1 gives heads.

Let $B$ be the event that coin2 gives heads.

Let $R$ be the event that both coins give heads.

Then $A$ and $B$ are independent, and are both depending on $R$.


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Example: Let $R$ take values from $00$ to $99$ with uniform probability. Let $A$ be the first decimal digit of $R$, and $B$ the last digit.