# Dirac Delta Evaluation of Function

I am struggling to understand what does the following expression evaluate to.

"I am looking for general answer, not actual evaluation - i.e is dirac dealta making the integral center around h(x)? what exatly is happeneing when dirac delta is being introduced"

$$\int g(\mathbf{x})p(x_1,y_1)\delta(x_1 - h(\mathbf x))dx_1dy_1$$

It has been taken from the following context: $${x}_{1}= \textrm{argmax} \ \mathcal{\alpha} (\mathbf{x}; I_{0})$$ $${x}_{2}= \textrm{argmax} \int \mathcal{\alpha} (\mathbf{x}; I_{1})p(y_{1}|x_{1},I_0)p(x_{1}|I_0)dx_1dy_1$$ where, $$p(y_{1}|x_{1},I_0)$$ is predictive distribution of the GP

$$p(x_{1}|I_0)$$ = $$\delta (x_{1}-\textrm{argmax} \ \mathcal{\alpha} (\mathbf{x}; I_{0}))$$

• The diract delta $\delta(x)$ has the property that, $$\int f(x)\delta(x)dx = f(0).$$ So, if we translate the delta by $x_0$, we will get, $$\int f(x)\delta(x-x_0)dx = \int f(\tilde{x}+x_0)\delta(\tilde{x})d\tilde{x} = f(x_0)$$ – TSF Nov 19 at 14:30
• I was able to follow that, however in the above case isnt it: $$\int g(\mathbf{x})p(x_1,y_1)\delta(0)dx_1dy_1 ?$$ – GENIVI-LEARNER Nov 19 at 14:39
• Also what does x~ in your equation represents? – GENIVI-LEARNER Nov 19 at 14:41
• $\tilde{x}$ is just a new variable, I didn't want to call it $x$ again. – TSF Nov 19 at 15:32