Continuous random variable probability density function

Question

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

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

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

Answer 1

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.

Answer 2

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.

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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

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

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