# Continuous random variable probability density function

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

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

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

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.

• 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 $b-\frac{4|4-3|}{5} = 0$ – 5xum Jul 31 '17 at 8:45
• 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. – nicola Jul 31 '17 at 8:47
• Ah, I was sloppy reading the inequalities. You are right, of course. – 5xum Jul 31 '17 at 8:52
• +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. – drhab Jul 31 '17 at 9:12
• @drhab You are totally right. I made an edit. – nicola Jul 31 '17 at 9:17

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 $$ax + \frac{8}{15}$$

• If they are asking for a cumulative distribution function, shouldn't it be $F(2)$+$F(3)$? – user373534 Jul 31 '17 at 9:00
• @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. – 5xum Jul 31 '17 at 9:01
• 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 $(2, 3)$? – user373534 Jul 31 '17 at 9:05
• @user373534 What is a "non-numerical integral"? – 5xum Jul 31 '17 at 9:06
• Oh I was trying to remember the name for it. Indefinite integral, sorry. Math isn't exactly on point today... – user373534 Jul 31 '17 at 9:09