# distribution questions

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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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
In the first line of this posting, there was an odd $\TeX$ anomaly: It said $b-a$, but it looked like $b$-$a$, as if there were a hyphen and no spacing, rather than a minus sign with normal spacing. I clicked on "edit" and it looked like b-a, which $\TeX$ would normally render as $b-a$. So I just typed a hyphen where there appeared to be a hyphen already, and then it looked normal. Might some non-standard character have been used there? –  Michael Hardy Nov 8 '12 at 20:07

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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