Value of a generalised probabilty density function with a dirac delta function at a single point

Question

I thought I understood the dirac delta function until I came across a question where I needed to evaluate the probability density function at a single point.

Probability density function: $$f_X(x) = \frac{1}{2}\delta(x-1) + \frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$

I would like to find the value of the function at $x=1$.

I know that the dirac delta function is used to account for the discontinuities in the cumulative distribution function, and that the dirac delta function is defined as: $$\delta(x) = \begin{cases} \infty,& x = 0\\ 0,& otherwise.\end{cases} $$

I could choose to ignore the discontinuity at $x=1$, but that could result in $$f_X(x=1) = \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}},$$ $$f_X(x=1) = \frac{1}{2} + \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}}$$

or any value in between.

But, according to the definition of the dirac delta function, at point $x=1$ the function could also evaluate to

$$f_X(x=1) = \infty + \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}} = \infty$$

This could be reasonable because the dirac delta function is not a traditional function, therefore I could accept a definition, if needed. The problem arises when I try to calculate conditional probabilities in any intervals containing $x=1$. For e.g., if I accept that $f_X(x=1) = \infty$, then

$$P(X=1|X \geq 1) = \frac{P(X=1)}{\int_{1}^{+\infty}(\frac{1}{2}\delta(x-1) + \frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{2}})} = \frac{P(X=1)}{\frac{1}{2\sqrt{2\pi}} \int_{-\infty}^{1} e^{-\frac{x^2}{2}}} = \infty,$$

which is clearly nonsensical. So then the questions arise, what is the value of $f_X(x=1)$ and why?

Answer 1

I would like to find the value of the function at x=1.

Your $f(x)$ is a "mixed density", it is not exactly a density function because your random variable is not absolute continuous. It has a positive probability mass in $X=1$ so you have

$$\mathbb{P}[X=1]=\frac{1}{2}$$

For the rest of the support you have a Standard Gaussian density thus, for example:

$$\mathbb{P}[X\geq 1]=\frac{1}{2}+\frac{1}{2}\left[1-\Phi(1)\right]\approx0.58$$

So for your question you will have

$$\mathbb{P}[X=1|X\geq 1]\approx\frac{50}{58}$$

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First observation

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Here is the correct (better, here is what your text means as) definition of your Dirac Delta

$$\delta(x-1)=\begin{cases} 1, & \text{if $x=1$ } \\ 0, & \text{elsewhere} \end{cases}$$

For detail, read here in the paragraph "Generalizations"

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Second observation

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In these cases it is better to define the random variable with its CDF that always exists rather then a "density" that will not exist.

Anyway I know that in some engineering applications, say telecommunication or Signal's Analysis they often use mixed density with impulse (Dirac delta), Rectangular (uniform) response et cetera et cetera

Answer 2

The Dirac delta is not a function. It is not defined by its values. Otherwise there would be no difference between $\delta (x) $ and $5 \delta (x) $. The Dirac delta is a generalized distribution and does not have values at all. The definition you give for it is incorrect; there is a way to twist and interpret it into a true statement, but it is very misleading at best. A generalized distribution is defined by its integrals. The probability distribution you gave is not a function and does not have a value at 1 at all. Because of the $\delta$, the entire expression is a generalized distribution and not a function.