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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?

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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 $\alpha=1/2, \beta=1/18$.

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  • $\begingroup$ What do you mean by the boundary points, as in x = 1 is the median? $\endgroup$ May 6, 2019 at 10:18
  • $\begingroup$ 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. $\endgroup$
    – gunes
    May 6, 2019 at 10:23
  • $\begingroup$ Could they not be continuous though? So you can't have two possible values at x = 1? $\endgroup$ May 6, 2019 at 10:27
  • $\begingroup$ 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. $\endgroup$
    – gunes
    May 6, 2019 at 10:29

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