This is for a homework question in which I am trying to find the $E(Y_n = min\{X_1,...,X_n\}$).

So far I have found that the minimum cdf, as below.

The minimum of $X_i$ is when all of $X_i > x$.

$Pr(Y_n \leq y) = Pr(X_i > x) = \prod_{i=1}^{n} Pr(X_i \leq 1-x) = F_X(1- x)^n $

Then I found the pdf like so:

$$ F_Y(y) = Pr(X_i > x) = \left\{ \begin{eqnarray} 0 \textrm{ for } y \leq 0 \\ (1-x)^n \textrm{ for } y \in (0,1) \\ 1 \textrm{ for } y \geq 1 \end{eqnarray} \right. $$

$$\frac{d}{dy}F_Y(y) = \frac{d}{dy}(1-y)^n = n(1-y)^{n-1} = f_Y(y)$$

... and then I tried to integrate for $ y \in (0,1)$ given the common definition of $E(Y) = \int_0^1 y f_Y(y) dy$

$$ \begin{aligned} E(Y_n) &= \int_0^1 yn(1-y)^{n-1}dy\\ &= n\int_0^1(1-u)(u)^{n-1} -du && \textrm{where } &&&u=1-y\\ &&& \textrm{where }&&&du=-dy\\ &= n\int_0^1(u-1)(u)^{n-1}du\\ &= n\int_0^1u^{n}-u^{n-1}du\\ &= n\left(\left[\frac{u^{n+1}}{n+1}\right]_0^1-\left[\frac{u^{n}}{n}\right]_0^1\right)\\ &= n\left(\left[\frac{(1-y)^{n+1}}{n+1}\right]_0^1-\left[\frac{(1-y)^{n}}{n}\right]_0^1\right)\\ &= n\left(\left[ 0 - \frac{1^{n+1}}{n+1}\right]-\left[0 - \frac{1^{n}}{n}\right]\right)\\ &= n\left(-\left[\frac{1^{n+1}}{n+1}\right]+\left[\frac{1^{n}}{n}\right]\right)\\ E(Y_n) &= 1^{n} - \frac{1^{n+1}}{n} \end{aligned} $$

However, this answer is apparently wrong. One source says that the correct answer is $$E(Y_n) = \frac{1}{n+1}$$

Is anyone able to steer me in the right direction in my integration? I haven't done calculus, let alone integration by u substitution, in 7 years or so! Thank you.

  • 1
    $\begingroup$ The very last line of your derivation is wrong when you multiply what's in the brackets by $n$. And for goodness sake simplify $1^n$ to $1$ $\endgroup$ – Hugh Nov 26 '16 at 23:52

After Hugh's advice, I reevaluated the integral and I just had an algebra problem. Below, find my answer. Note, I change the $1^{n}$ and $1^{n+1}$ to $1$, for goodness' sake.

$$ \begin{aligned} E(Y_n) &= n\left(-\left[\frac{1}{n+1}\right]+\left[\frac{1}{n}\right]\right)\\ &= \frac{n}{n}-\frac{n}{n+1}\\ &= \frac{n+1}{n+1}-\frac{n}{n+1} &&\textrm{since } \frac{n}{n} = \frac{n+1}{n+1}=1\\ E(Y_n) &= \frac{1}{n+1} \end{aligned} $$

Thank you, Hugh!


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