If Mean Squared Error = Variance + Bias^2. Then How can the Mean Squared Error be lower than the Variance I was reading the Introduction to Statistical Learning. Here it is shown that:- 
In a later example, the train and test MSE are plotted. I wanted to know if both the bias^2 and variance are positive quantities then how can MSE be lower than the Variance.

 A: That's called overfitting. The apparent MSE on the training data is lower than the variance, but this was only achieved by making a model overly complicated so that it could follow random fluctuations art individual data points ("chasing noise"). Once you try to predict on new data MSE is much worse. I.e. the real MSE of predictions from the model is not lower than the variance. 
A: The formula reproduced in the question is exact and hence not compatible with an "MSE lower than the Variance". When you mention one observes an "MSE lower than the Variance" on the provided graph (assuming the minimum MSE is the model variance), it is because you consider empirical MSE and variances, rather than the theoretical quantities, which are expectations against the model distribution.
A: You seem to think that there is a case showing the variance being larger than the MSE, but it is far from clear how you are seeing that. In machine learning, Y is modeled as being equal to some function of X, plus a random error term. That error is, as it is in this example, often represented with an epsilon, $\epsilon$. In this model, an estimator function equal to the "real" dependency of Y on X will have an MSE equal to the variance of $\epsilon$. An estimator other than the "real" dependency will have an MSE equal to the variance of $\epsilon$, plus the variance between the "real" dependency and the estimator used. Thus, the MSE of the estimator will be greater than or equal to the variance $\underline{\text{of }}\underline{\epsilon}$. It can be, and any decent estimator will be, less than the variance $\underline{\text{of Y}}$. If the MSE of an estimator were greater than the variance of Y, then ignoring X completely and just predicting that Y will be equal to the mean of Y would be a better estimator.
