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#1
11-14-2017, 03:39 PM
 Nagsinde2002 SOA Join Date: May 2017 College: University of Wisconsin - Madison (1st year) Posts: 12
Help Needed with these Problems

I am stuck on how to solve these problems. If someone can help me, that would be greatly appreciated.

1) Suppose that X and Y are independent continuous random variables
with density functions fX and fY . Let T = min(X, Y ) and V = max(X, Y ).
Find the marginal density functions fT of T and fV of V.

I know that I should find the CDF of T and V and differentiate those to give me my marginal density functions, but how do I calculate the CDF of T and V?

2) Let 0 < p < 1 and 0 < r < 1 with p not being equal to r. You repeat a trial with success probability p until you see the first success. I repeat a trial with success probability r until I see the first success. All the trials are independent of each other.

a)What is the probability that you and I performed the same number of trials?
b) Let Z be the total number of trials you and I performed altogether. Find the possible values and the probability mass function of Z.

3) Let X and Y be independent exponential random variables with
parameter 1.

(a) Calculate the probability P(Y ≥ X ≥ 2).
(b) Find the density function of the random variable X − Y .

Sorry for this long post, but hopefully someone can help me with these problems.
#2
11-14-2017, 03:41 PM
 NormalDan Member CAS Join Date: Dec 2016 Location: NJ Posts: 4,967

Why not show what you've got thus far with your specific questions. People won't just solve them for you.
#3
11-17-2017, 03:47 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 336

These look too hard for exam P. Where did you find them?

I will do #2 for you.

Let X be the number of trials until first success (w/ success probability p)
Let Y be the number of trials until first success (w/ success probability r)

(a)
\begin{align*}\sum_{i=1}^{\infty}P(X=i\text{ }\cap\text{ }Y=i)&=\sum_{i=1}^{\infty}P(X=i)P(Y=i) &&[\text{independence}]\\&=\sum_{i=1}^{\infty} (1-p)^{i-1}p(1-r)^{i-1}r\\
&=pr\sum_{i=1}^{\infty} (1-p)^{i-1}(1-r)^{i-1}\\
&=pr\sum_{i=0}^{\infty} (1-p)^{i}(1-r)^{i}\\
&=pr\sum_{i=0}^{\infty} ((1-p)(1-r))^{i}\\
&=pr\cdot \dfrac{1}{1-(1-p)(1-r)}\\
&=\dfrac{pr}{p+r-pr}
\end{align*}

(b) Let $z \geq 2$ be an integer.
\begin{align*}P(X+Y=z)=\sum_{i=1}^{z-1}P(X=i\text{ }\cap\text{ }Y=z-i)&=\sum_{i=1}^{z-1}P(X=i)P(Y=z-i) &&[\text{independence}]\\
&=\sum_{i=1}^{z-1}(1-p)^{i-1}p(1-r)^{z-i-1}r\\
&\vdots \\
&=\frac{pr\left((1-r)^{z}(1-p)+(1-p)^{z}(r-1)\right)}{(p-1) (r-1) (p-r)}
\end{align*}

unless I made a mistake somewhere...
(PMF is 0 if z is not an integer >=2)
#4
11-17-2017, 04:04 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,076

3a is just a double integral. Draw a picture. For 3b. here is a more general case:

http://www.math.wm.edu/~leemis/chart...ialLaplace.pdf

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