Integration by Parts with Probability Density Functions

Question

For $Y$ a random variable that has sample space $S = [0, \infty)$ with a density function, $$ f_Y (y) = 9ye^{−3y} $$

Find:

1) $P\{Y > \frac 1 3\}$

2) $\operatorname{E}\left(\frac 1 Y \right)$

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I learned how to integrate by parts (years ago, so I'm a bit rusty) and I know there's countless resources online, but in terms of probability density functions I get a bit confused. My notes from lecture uses substitution when simplifying an integral something with gamma or E? I'm guessing this shortcut in simplifying is how to get through this problem.

Answer 1

For the first integral, you have

$$P(Y > \frac{1}{3}) = \int_{1/3}^{\infty} 9ye^{-3y} \> dy$$

Integration by parts is loosely just

$$\int u \> dv = uv - \int v \> du$$

We see that letting $u=y$ will simplify computation. This leaves

$$P(Y > \frac{1}{3}) = 9 \left( \left. (y(-\frac{1}{3}e^{-3y}))\right|_{1/3}^{\infty} - \int_{1/3}^{\infty} -\frac{1}{3}e^{-3y} \> dy \right)$$

You should be able to evaluate from here. Using LOTUS the expectation is just

$$E(\frac{1}{y}) = \int_0^{\infty} \frac{1}{y} 9ye^{-3y} \> dy = \int_0^{\infty} 9e^{-3y} \> dy$$

Answer 2

For 1) compute

$$\int_{1/3}^{\infty} 9ye^{-3y} dy= \lim_{t\to\infty} \int_{1/3}^{t} 9ye^{-3y}dy$$

Using a substitution will make this less tedius, but not by much. To skip that, first take out the $9$, then use $u=y$, $dv=e^{-3y}dy$ in the integration by parts formula:

$$\int_a^b u \ dv = [u(y)v(y)]\Big|_a^b - \int_a^b v \ du$$

For $2$, no parts necessary.

$$E(g(Y)) =\int_0^{\infty} g(y) f_Y(y) dy$$

You may be able to see quickly that the $y's$ cancel for your $g(y)$.

Answer 3

\begin{align} \int 9y e^{-3y} \, dy = 9 \int y\Big(e^{-3y} \, dy\Big) = 9 \int y\,dv = 9\left( yv - \int v\,dy \right) \end{align} $$ dv = e^{-3y} \, dy, \qquad v = -\frac {e^{-3y}} 3 $$ and so on. In the expression $\left[ yv \vphantom{\dfrac 1 1} \right]_{y=1/3}^{\infty}$ you can use L'Hopital's rule.