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  1. Let $X$ have a uniform distribution $\operatorname{U}(0,1)$, and let $Y = a + (b-a)X,\, a < b$. (a) Find the distribution function of $Y$. (b) How is $Y$ distributed?

  2. Let $X_1$ , $X_2$ be independent random variables representing lifetimes (in hours) of two key components of a device that fails when and only when both components fail. Say each $X_i$ has an exponential distribution with mean 1000. Let $Y_1 = \min(X_1, X_2)$ and $Y_2 = \max(X_1, X_2)$, so that the space of $Y_1$ and $Y_2$ is $0 < y_1 < y_2 < \infty$. (a) Find $P(Y_1 = y_1, Y_2 = y_2)$ (b) Compute the probability that the device fails after 1200 hours.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Did Nov 8 '12 at 19:02
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Please note that your post consists of two unrelated questions. In general, it is better to put unrelated questions in separate posts. – André Nicolas Nov 8 '12 at 19:40

1. We find $\Pr(Y\le y)$. This is $\Pr(a+(b-a)X)\le y$, which is $\Pr\left(X\le \frac{y-a}{b-a}\right)$.

If $y-a\le 0$, that is, if $y\le a$, this probability is $0$.

If $\dfrac{y-a}{b-a}\ge 1$, that is, if $y\ge b$, this probability is $1$.

And finally, the interesting part. If $a\lt y\lt b$, this probability is $\dfrac{y-a}{b-a}$.

The three sentences above describe the cumulative distribution function of $Y$. For the density, differentiate. We get density $\dfrac{1}{b-a}$ on $(a,b)$ and $0$ elsewhere. Note that $Y$ has uniform distribution on $(a,b)$ (or equivalently, $[a,b]$).

2. a) Our distributions are continuous. So, as the problem is currently stated, the probability is $0$.

For b), each of $X_1$ and $X_2$ has density function of the shape $e^{-t/1000}$ when $t\gt 0$. So the probability the first is alive after $1200$ hours, by integration, is $e^{-1200/1000}$. The same is true for the second.

Because $X_1$ and $X_2$ are independent, the probability both components are still alive after $1200$ hours is the product of the individual probabilities, that is, $e^{-2400/1000}$. If we interpret "device fails after $1200$ hours" as meaning that the lifetime of the device is at least $1200$, that gives the answer to the question.

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