Is it possible to find dependence between $X$ and $Y$ from distribution of $X-Y$? Consider $X,Y \sim \mathcal N(0,1)$, I am exploring all the possible ways to find dependence. Correlation, Eye Balling from scatter plot are I think obvious methods.
I am trying to think in a different direction just to have a better understanding. So, When $Y$ is dependent on $X$ what would $X-Y$ distribution look like?
From what I am aware of $X-Y \sim \mathcal N(0,2)$ when $X$ and $Y$ are independent. 
Consider a distribution show in the graph below, I want to comment on dependence just from the graph dist
If this is not possible how can joint probability tell be about dependence.
 A: $\newcommand{\v}{\operatorname{var}}$You have $X,Y \sim N(0,1).$
You did not say anything about their joint distribution beyond that --- in particular whether they are jointly normally distributed.
If they are independent, then they are jointly normally distributed. In that case you have $\v(X-Y) = \v(X)+\v(Y) = 2,$ so $\operatorname{sd}(X-Y) = \sqrt 2.$
Suppose they are jointly normal but not independent, and the correlation is $\rho.$ Then
\begin{align}
\v(X-Y) & = \v(X)+\v(Y) - 2\operatorname{cov}(X,Y) \\
& = \v(X) + \v(Y) - 2\rho\sqrt{\v(X)} \sqrt{\v(Y) } \\
& = 1 + 1 -2\rho.
\end{align}
Since $-1\le\rho\le1,$ we have $0 \le \v(X-Y) \le 4.$
So $X-Y\sim N(0,2-2\rho).$ Knowing the variance of $X-Y,$ you can find $\rho,$ and knowing $\rho$ you can find the variance of $X-Y.$
The distributions in the graph do not look normal except that the red one may be a crude approximation. Perhaps density estimation based on a small sample from a normally distributed population could have produced them.
It took me a while (20 seconds?) to suspect that the hyphens in the graphic might have been intended to be minus signs.
\begin{align}
\text{minus sign: } & X-Y \\[10pt]
\text{hyphen: } & X \text{ - } Y \\[10pt]
\text{hyphen: } & X \text{-} Y
\end{align}
