Need help with a practice probability question for Exam P!

#1
Here is the question - The length of time, in years, that a person will remember an actuarial statistic is modeled by an exponential distribution with mean 1/Y. In a certain population, Y has a gamma distribution with a=2 and B=1/2. Calculate the probability that a person drawn at random from this population will remember an actuarial statistic less than 1/2 year.

Here's how far I have gotten, which isn't very btw.
T= length someone remembers an actuarial statistic
T|Y has an exp distribution with mean 1/Y so f(t|y)=ye^(-ty)
f(y)=4ye^(-2y)
We need to figure out P(T<1/2).
I figure I could find f(t) by multiplying f(t|y) and f(y) to get f(t,y) and then integrating that with respect to y to get f(t). But the integral looks too messy.

Please explain this to me without the use of MGF. The manual I have explains it using MGF which isn't making any sense to me.

Thanks in advance!
 

Dason

Ambassador to the humans
#2
When you look at f(t|y)f(y) is there any way to get that to look like a familiar distribution with respect to y? Possibly one of the distributions you're working with? You might need a few constants but see if you can find anything that looks like a distribution you know.

You'll probably use this kind of trick *a lot* in the future so you might want to get used to spotting the kernel of a distribution in an integral which makes evaluating the integral a lot easier.
 

BGM

TS Contributor
#3
Try to make yourself familiar with the gamma distribution and its integral:

If \( X \sim Gamma(\alpha, \beta) \), then its p.d.f.
\( f_X(x) = \frac {1} {\Gamma(\alpha)\beta^\alpha} x^{\alpha-1}
e^{-\frac {x} {\beta}}, x > 0, \alpha > 0, \beta > 0 \)
\( \Rightarrow \int_0^{+\infty} \frac {1} {\Gamma(\alpha)\beta^\alpha}
x^{\alpha-1} e^{-\frac {x} {\beta}} dx = 1 \)
\( \Rightarrow \int_0^{+\infty} x^{\alpha-1} e^{-\frac {x} {\beta}} dx
= \Gamma(\alpha)\beta^\alpha \)

Now \( f_{T,Y}(t, y) = f_{T|Y=y}(t|y)f_Y(y) = ye^{-ty}4ye^{-2y}
= 4y^2e^{-(t+2)y} \)
You should be able to figure out the corresponding parameters
and get the p.d.f. for \( T \).

Actually you can use
\( Pr\{T < \frac {1} {2}\}
= \int_0^{+\infty} Pr\{T < \frac {1} {2}|Y = y\}f_Y(y)dy \)
so you just need to find out the c.d.f. of \( T|Y = y \),
which is easy as it is just an exponential distribution