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This is a problem on the joint distribution of a discrete and a continuous random variable.

Kitty Oil Co. has decided to drill for oil in 10 different locations; the cost of drilling at each location is 10,000. (Total cost is then 100,000.)

The probability of finding oil in a given location is only 0.2, but if oil is found at a given location, then the amount of money the company will get selling oil (excluding the initial 10,000 cost) from that location is an exponential random variable with mean 50,000.

Let Y be the random variable that denotes the number of locations where oil is found, and let Z denote the total amount of money received from selling oil from the locations.


Here what is expectation of $Z$ and $\mathbb{P}(Z>10000|Y=1)$?

How should one evaluate such questions in general??

Thanks in advance

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It is not clear whether $Z$ is income or net income. No big deal, if we can handle one we can handle the other. We use the gross income interpretation.

Let $Z_1,Z_2,\dots, Z_{10}$ be the amount of money made from digs $1,2,\dots,10$. Then $Z=Z_1+Z_2+\cdots+Z_{10}$. By the linearity of expectation, we have $E(Z)=E(Z_1)+\cdots +E(Z_{10})=10E(Z_1)$.

To find $E(Z_1)$, note that $Z_1=0$ with probability $1-p$, where $p$ is the probability of finding oil if one digs, currently unreadable. And given that the well was successful, the expectation is $50000$. Thus $E(Z_1)=(1-p)(0)+(p)(50000)$.

For the probability that $Z\gt 10000$ given $Y=1$, we just want the probability that an exponential with mean $50000$ is greater than $10000$.

Remark: If we interpret $Z$ as net income, for the expectation question subtract $100000$.

For the probability question, find the probability that an exponential with mean $50000$ is greater than $110000$.

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  • $\begingroup$ in that case, as Z is the sum of exponential random variables, does Z becomes a gamma(10,\frac{1}{50000})? and what is the relation between Z and Y? $\endgroup$ – tap1cse Sep 26 '13 at 9:19
  • $\begingroup$ For the first question, yes, but we don't need to observe that to get the mean. As for relationship between $Z$ and $Y$, I imagine we could describe the joint distribution in some fairly nice way. Certainly can nicely describe the conditional distribution of $Z$ given $Y$, gamma with parameters $y$, $\lambda$. $\endgroup$ – André Nicolas Sep 26 '13 at 9:53
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    $\begingroup$ You are welcome. When computing expectations, we quite frequently want to (and can) bypass finding the distribution, which can be hard to get at and use. That's why linearity of expectation, also using conditional expectation, are such nice tools. $\endgroup$ – André Nicolas Sep 26 '13 at 15:46

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