-1
$\begingroup$

$f(x) = \begin{cases} \frac{4x^3}{15}, & 0\leq x\leq1, \\ ax+\frac{8}{15}, & 1\leq x\lt 2, \\ b-\frac{4|x-3|}{5}, & 2\leq x \lt 4 \\ 0, &\text{otherwise} \end{cases}$

How do you find the value of $a$ or $b$ here? I am aware that $P(a \leq X \leq b) = \int_a^bf(x)dx$, but how do I use this to solve for either variable?

Also would the cumulative distribution function of $f(x)$ for $2\leq x\lt3$ just be

$F(x) =\int_2^2ax+\frac{8}{15}dx+\int_2^3b-\frac{4|x-3|}{5}dx,\\$

$\endgroup$
3
$\begingroup$

You have two parameters ($a$ and $b$), so you need two conditions. The first one comes from the fact that for $x=1$ you must have:

$$ \frac{4}{15} = a + \frac{8}{15} $$ because the value $1$ belongs to both first and second case of the $f$ definition.

The second condition comes from the fact that $f$ is a probability density function and, as such, the integral of it in its domain must be 1:

$$ \int_0^4 f(x) dx = 1 $$ So you have two unknowns and two equations.

$\endgroup$
  • $\begingroup$ There is no need for $f$ to be continuous at $x=1$, even if $f$ is discontinuous, $F$ is still continuous. Also, if you demand continuity at $x=1$, why not demand continuity at $x=4$? That would give the equation $b-\frac{4|4-3|}{5} = 0$ $\endgroup$ – 5xum Jul 31 '17 at 8:45
  • 3
    $\begingroup$ I do not demand continuity on $x=1$. See the definition of $f$. How do you calculate $f(1)$? Which case do you select? Since $f(1)$ cannot be two values, that equality must hold. $\endgroup$ – nicola Jul 31 '17 at 8:47
  • 1
    $\begingroup$ Ah, I was sloppy reading the inequalities. You are right, of course. $\endgroup$ – 5xum Jul 31 '17 at 8:52
  • 1
    $\begingroup$ +1 But I wouldn't call $f$ a "distribution function". If that term is used then it is for the CDF. Here $f$ is a PDF (probability density function) as stated in the title of the question. $\endgroup$ – drhab Jul 31 '17 at 9:12
  • $\begingroup$ @drhab You are totally right. I made an edit. $\endgroup$ – nicola Jul 31 '17 at 9:17
2
$\begingroup$

How do you find the value of $a$ or $b$ here? I am aware that $P(a \leq X \leq b) = \int_a^bf(x)dx$, but how do I use this to solve for either variable?

First of all, a huge warning: it's very dangerous to write $P(a\leq X\leq b)$ in this case, because you are overloading the variables $a,b$. The letters $a$ and $b$ are already used as parameters in the expression $f(x)$, so using them for the bounds of integration can cause confusion at best, and plain our incorrect statements at worst.


Also would the cumulative distribution function of $f(x)$ for $2\leq x\lt3$ just be $F(x) =\int_2^2ax+\frac{8}{15}dx+\int_2^3b-\frac{4|x-3|}{5}dx$

Again, huge warning about notation: what you wrote on the right is an expression independent of $x$, so there shouldn't be $F(x)$ on the right. Instead, what you are looking for is

$$F(3)-F(2)$$

which is equal to $$\int_{2}^3 f(x)dx$$

Now, because $f(x)=b-\frac{4|x-3|}{5}$ for $x\in[2,3]$, you can replace that with

$$\int_2^3\left(b-\frac{4|x-3|}{5}\right)dx$$


And finally, to your original question:

You know that if $f$ is a probabilistic distribution, then

$$\int_{-\infty}^\infty f(x)dx=1$$

which in your case means that $$\int_0^4f(x)dx=1$$

and this can give you one equation for $a,b$.

The other equation comes from the fact that in your definition, for $x=1$, $f(x)$ must at the same time be equal to $$\frac{4x^3}{15}$$ and $$ax + \frac{8}{15}$$

$\endgroup$
  • $\begingroup$ If they are asking for a cumulative distribution function, shouldn't it be $F(2)$+$F(3)$? $\endgroup$ – user373534 Jul 31 '17 at 9:00
  • 1
    $\begingroup$ @user373534 No. It's $F(3) - F(2)$ because $F(3)$ is "what accumulates on $(-\infty, 3)$", and $F(2)$ is "what accumulates on $(-\infty, 2)$. The difference, then, is "what accumulates on $(2,3)$. And no offense, but if you don't know that, you should re-read the entire chapter on cumulative distribution functions. $\endgroup$ – 5xum Jul 31 '17 at 9:01
  • $\begingroup$ Ah yes it makes sense now. But wouldn't the cumulative distribution function be represented as a piecewise function, similar to that of $f(x)$? Why is it not a non-numerical integral with the range $(2, 3)$? $\endgroup$ – user373534 Jul 31 '17 at 9:05
  • $\begingroup$ @user373534 What is a "non-numerical integral"? $\endgroup$ – 5xum Jul 31 '17 at 9:06
  • $\begingroup$ Oh I was trying to remember the name for it. Indefinite integral, sorry. Math isn't exactly on point today... $\endgroup$ – user373534 Jul 31 '17 at 9:09

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.