2
$\begingroup$

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.

$\endgroup$
  • 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
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.