0
$\begingroup$

I'm trying to study for an exam and I'm having a bit of difficulty understanding the given solution to a particular problem. Given $F_X(x)$ it asks the reader to find $f_X(x)$

$$ F_X(x) = \begin{cases} 0 & x < -5 \\ \frac{x+5}{8} & -5\leq x < -3 \\ \frac{1}{4} & -3\leq x < 3 \\ \frac{1}{2} & 3\leq x < 4 \\ \frac{1}{2} + \frac{x-4}{2} & 4\leq x < 5 \\ 1 & x \geq 5 \end{cases} \ $$

The provided answer is in terms of the Dirac Delta function $\delta(x)$ and the unit step function $U(x)$

$$f_X(x) = \frac{1}{8}(U(x+5) - U(x+3)) + \frac{1}{4}\delta(x-3) + \frac{1}{2}(U(x-4) - U(x-5)), x \in \mathbb{R} $$

I would appreciate if somebody could give me an explination for this answer. Thank you!

$\endgroup$
0
$\begingroup$

When the CDF of $X$ has a jump at $x_0$, i.e. $F(x_0^-) \neq F(x_0)$, $X$ will take on $x_0$ with probability $F(x_0)-F(x_0^-)$. This can be represented in a density as $(F(x_0) - F(x_0^-)) \delta(x-x_0)$. This is just the fact that the derivative of a step function is a dirac delta, or if you just integrated that term over any set containing $x_0$, it would be $F(x_0) - F(x_0^-)$ (i.e. the probability of getting exactly $x_0$), and zero for any set not containing $x_0$.

If this confuses you, try $F_X(x) = (1-p) U(x) + p U(x-1)$. This is the CDF of a Bernoulli(p) R.V.

$\endgroup$
0
$\begingroup$

Another (easier, I guess) way of looking at it: it's a mixed discrete-continuous rv (I hope this term is not very incorrect). By taking derivatives of $F_X(x)$ you can find that for $-5 \leq x <3 \ f_X(x) = \frac{1}{8}$ and for $4 \leq x <5 \ f_X(x)=\frac{1}{2}$, these two being unit step functions. Also there's a jump at $x=3$ where $f_X(x) = \frac{1}{4}$, this is a Dirac delta function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.