# Probability density functions

The probability density function of $$X$$ is defined by:

\begin{align} f(x) = \begin{cases} \alpha & \quad, 0 \le x \le 1 \\ \beta(x-4)^2 & \quad, 1 \le x \le 4 \end{cases} \end{align}

Show that the exact values of $$\alpha$$ and $$\beta$$ are $$\frac{1}{2}$$ and $$\frac{1}{18}$$ respectively:

We know that the total area under the probability density function is equal to $$1$$. But using this fact only gives us $$\alpha + 9\beta = 1$$.

Are there any other conditions that would give us another simultaneous equation, or allow us to conclude either $$9\beta = 0.5$$ or $$\alpha = 0.5$$ or something similar?

For both intervals you have $$x=1$$ included, the two expressions for the two intervals must be equal at $$x=1$$. Left boundary is $$\alpha$$, and right boundary is $$\beta(x-4)^2|_{x=1}=9\beta$$. So, we have another equality: $$\alpha=9\beta$$. Solving for the two, i.e. $$\alpha=9\beta, \alpha+9\beta=1$$ leaves us with solution $$\alpha=1/2, \beta=1/18$$.
• No, I clarified the expression. Since both left interval and right interval intersect at $x=1$, the two functions defined in these intervals must yield the same value at their boundary, i.e. $x=1$. I mean what would you say if asked you $f(1)$? Is it $\alpha$ or $\beta(x-4)^2$? For the function to have only one value at $x=1$, the two must be equal. Commented May 6, 2019 at 10:23
• PDF may be discontinuous, but it is a function and it cannot have two values for a specific $x$. If your question was for $0\leq x < 1$ and $1\leq x \leq 4$, we couldn't say anything, because $x=1$ isn't included in the first interval. But, since it's included, both right and left functions must yield the same value. Commented May 6, 2019 at 10:29