Is there a possibility that there is no relationship between some inputs and outputs? I'm doing machine learning projects. I took a look at many datasets I worked with, mostly there are already famous datasets that everyone uses.

Let's say I decided to make my own dataset. Is there a possibility that my data are so random so that no relationship exists between my inputs and outputs? This is interesting because if this is possible, then no machine learning model will achieve to find an inputs outputs relationship in the data and will fail to solve the regression or classification problem.

Moreover, is it mathematically possible that some values have absolutely no relationship between them? In other words, there is no function (linear or nonlinear) that can map those inputs to the outputs.

Now, I thought about this problem and concluded that, if there is a possibility for this, then it will likely happen in regression because maybe the target outputs are in the same range and the same features values can correspond to the same output values and that will confuse the machine learning model.

Have you ever encountered this or a similar issue?
Of course, it's possible to define a problem where there is no relationship between input $x$ and output $y$. In general, if the mutual information between $x$ and $y$ is zero (i.e. $x$ and $y$ are statistically independent) then the best prediction you can do is independent of $x$. The task of machine learning is to learn a distribution $q(y|x)$ that is as close as possible to the real data generating distribution $p(y|x)$.

For example, looking at the common cross-entropy loss, we have 
$$
\begin{align}
H(p,q) = -\mathbb{E}_{y,x \sim p}\left[\log q(y|x)\right] &amp; = \mathbb{E}_{x\sim p}\left[\text{H}(p(y|x)) + \text{D}_{\text{KL}}(p(y|x)\|q(y|x))\right]  \\
&amp; = \text{H}(p(y)) + \mathbb{E}_{x \sim p}\left[\text{D}_{\text{KL}}(p(y)\|q(y|x))\right],
\end{align}
$$
where we have used the fact that $p(y|x)=p(y)$ since $y$ and $x$ are independent. From this, we can see that the optimal predicted distribution $q(y|x)$ is equal to $p(y)$, and actually independent of $x$. Also, the best loss you can get is equal to the entropy of $y$.