a) If f(y) is a probability density function, then both f(y) ≥ 0 for all y in its support, and the integral of f(y) over its entire support should be 1. eˣ > 0 for all real x, so the first condition is met. We have
[tex]\displaystyle \int_{-\infty}^\infty f(y) \, dy = \frac14 \int_0^\infty e^{-\frac y4} \, dy = -\left(\lim_{y\to\infty}e^{-\frac y4} - e^0\right) = \boxed{1}[/tex]
so both conditions are met and f(y) is indeed a PDF.
b) The probability P(Y > 4) is given by the integral,
[tex]\displaystyle \int_{-\infty}^4 f(y) \, dy = \frac14 \int_0^4 e^{-\frac y4} \, dy = -\left(e^{-1} - e^0\right) = \frac{e - 1}{e} \approx \boxed{0.632}[/tex]
c) The mean is given by the integral,
[tex]\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \int_0^\infty y e^{-\frac y4} \, dy[/tex]
Integrate by parts, with
[tex]u = y \implies du = dy[/tex]
[tex]dv = e^{-\frac y4} \, dy \implies v = -4 e^{-\frac y4}[/tex]
Then
[tex]\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \left(\left(\lim_{y\to\infty}\left(-4y e^{-\frac y4}\right) - \left(-4\cdot0\cdot e^0\right)\right) + 4 \int_0^\infty e^{-\frac y4} \, dy\right)[/tex]
[tex]\displaystyle \cdots = \int_0^\infty e^{-\frac y4} \, dy[/tex]
[tex]\displaystyle \cdots = -4 \left(\lim_{y\to\infty} e^{-\frac y4} - e^0\right) = \boxed{4}[/tex]
