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Let $X_n$ with $n \ge 1$ a sequence of random variables such that, for each $n$, $X_n$ has density $n(1-x)^{n-1}$ with $x \in (0,1)$ ($X_n$ is $Beta (1,n)$). For each $n\ge 1$, let $T_n = (1-X_n)^n$

I need to find the distribution of $T_n$ and calculate $P(T_1 + \dots + T_{108} >50)$

I'm a bit confused on how to work with this sequence. I tried:

$P(T_n \le x) = \int_{0}^{x} (1-n(1-x)^{n-1})^n dx$

But this doesn't seem to work. How should I work with sequences such as these?

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    $\begingroup$ The density of the function of a random variable is not simply that function of the density, but given by the change of variables rule, see, e.g., en.wikipedia.org/wiki/…. $\endgroup$ Commented Nov 1, 2016 at 12:46
  • $\begingroup$ I was actually trying to find the cdf to then derive it. Wouldn't it work even with the change in variable? $\endgroup$ Commented Nov 1, 2016 at 12:49
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    $\begingroup$ Another useful property should be a symmetry property of the Beta distribution to the transformation $1-X$, see the Wikipedia page. $\endgroup$ Commented Nov 1, 2016 at 12:50
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    $\begingroup$ Yes, having the cdf would solve the problem, but your approach in the originial post is not the way to go, $\endgroup$ Commented Nov 1, 2016 at 12:50

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Strong hint: If $X_n\sim \mathcal{Be}(1,n)$, the density of the distribution is$$f_n(x)=n(1-x)^{n-1}$$and its cdf is$$F_n(x)=1-(1-x)^n$$Therefore,$$T_n=1-F_n(X_n)$$Now,

if $X_n\sim F_n$, what is the distribution of $F_n(X_n)$?

Additional hint: the formula $$P(T_n \le x) = \int_{0}^{x} (1-n(1-x)^{n-1})^n dx$$ is not correct.

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  • $\begingroup$ Just checking, in this case $F_n(X_n)$ would be a regularized incomplete beta function right? Or I am missing something special here? I'll see if I can solve it with the hints! Thanks! $\endgroup$ Commented Nov 1, 2016 at 20:48
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    $\begingroup$ The important item is that $F_n$ is the cdf of $X_n$. $\endgroup$
    – Xi'an
    Commented Nov 1, 2016 at 20:52
  • $\begingroup$ Okay, I think I got. $T_n$ would have a distribution $f(t_n) = 1$, which is a uniform (0,1). Now for $P(T_1,\dots,T_{108} >50)$ I could use a normal approximation with mean $108/2$ and variance $108/12$. Thanks for all the help! $\endgroup$ Commented Nov 2, 2016 at 18:12
  • $\begingroup$ Almost there: the law of the sum of $k$ uniforms is called an Irwin-Hall distribution so you do not need to use the Normal approximation. $\endgroup$
    – Xi'an
    Commented Nov 2, 2016 at 18:15

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