Is Bias term required for RBF with Gaussian kernel? For standard logistic regression, we add a bias term (1) in the features. Is it required when RBF is used with Gaussian Kernel?
 A: I'm not totally clear on what you have in mind, so this answer may be a bit off base.
But if I understand you correctly, you're considering the RBF 
$$
\begin{align}
f(a,b) &= \exp(-\gamma(a-b)^\top(a-b)) \\
& = \exp(-\gamma \| z \|_2^2)
\end{align}
$$
and asking whether you should use 
$$
x=[x_1, x_2 , ..., x_n ]^\top \\
y=[y_1, y_2, ..., y_n ]^\top
$$
or
$$
\tilde{x} = [ 1, x_1, x_2, ..., x_n ]^\top \\
\tilde{y} = [ 1, y_1, y_2, ..., y_n ]^\top
$$
in evaluating $f$.
If this is the case, there's no difference, because the $1$s cancel. Another way to think about it is that kernel methods only care about the kernel function values, not the raw features; in this case, including the intercept among the features is redundant because the RBF kernel only varies with respect of the norm of the difference of the input vectors. 
That is, 
$$
\begin{align}
\| z \| ^ 2_2&=(\tilde{x}-\tilde{y})^T(\tilde{x}-\tilde{y}) \\
&= \left\| ~ [1 - 1, x_1 - y_1, x_2 - y_2, \cdots, x_n - y_n]^\top ~ \right\|_2^2 \\
&= 0 + (x_1 - y_1)^2 + (x_2 - y_2)^2 + \cdots + (x_n - y_n)^2 \\
&=(x-y)^T(x-y).
\end{align}
$$ Because $f$ only varies through $z$, appending $1$ to the vectors makes no difference, since it cancels out when taking the inner product of the difference.
