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I am having a dreadful time making progress with the following problem, and would greatly appreciate any pointers you had.

The problem states that a device's risk of failure within 8 hours of activation, X, varies from day to day according to the probability density function $f(x) = 20(1-x)/9$ for $0.1 < x < 1$, and $f(x) = 20x$ for $0 < x < 0.1$. I need to find both the conditional and the marginal probabilities of exactly 3 failures within 8 hours of activation among the 4 devices activated today. I am to assume that the number of failures follows a binomial distribution.

With other problems, we are given f as a function of both x and y, and it is usually straightforward to find the conditional pdf as $h(y|x) = f(x,y) / f_1(x)$ and the marginal distribution as $f_1(x) = \int_{\infty}^{\infty} f(x,y) dy$.

I assume I need to do something along these lines, but obtaining $f(x,y)$ from my specific information is giving me a headache. Instead of x in the equation above, I can put ${\rm Binomial}(x,n,p)$ but, since I know $x$ and $n$, I still only have one variable - $p$.

I would be very grateful to anyone who could get me started with this, please. Thank you in advance.

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    $\begingroup$ Can you make your problem clearer? What does $x$ and $y$ refer to? $\endgroup$
    – emrea
    Jun 13, 2012 at 21:20
  • $\begingroup$ I see no y in the specific problem but you haven't explained what x is. Is it time? Is f a failure rate as a function of time? $\endgroup$ Jun 13, 2012 at 21:37
  • $\begingroup$ Thanks, Michael. The population parameter X is the risk of failure. What's confused me, I'm sure, is that the variable in the probability density function seems itself to be a probability. The exact wording of the scenario is as follows: "The risk of failure within 8 hours of activation, X, varies from day to day according to the probability density function f(x) = [see the original post]. In an audit of device performance, a day is chosen at random by auditors to review device outcomes." $\endgroup$
    – Craig ward
    Jun 14, 2012 at 10:40

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I think the statement of the problem assumes that the number of failures $N$ has a conditional binomial distribution $\text{Bin}(4, \theta)$ given $X=\theta$, where $X$ is the random variable whose law is defined in your text (notations are not very convenient !) In order to calculate the marginal probabilities $\Pr(N=n)$ you have to use the "marginalization" formula $\Pr(N=n) = \int \Pr(N=n \mid X=\theta)f(\theta)\mathrm{d}\theta$.

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