<p>I guess <span class="math-container">$m &lt; 1$</span>. Otherwise <span class="math-container">$x^{1/(m-1)} \to \infty$</span> as <span class="math-container">$x \to 0$</span>.</p>
<p>There's a possibility that I made a mistake, so let me know if you spot one.</p>
<p>Let's first compute the cumulative distribution function of <span class="math-container">$X_2$</span>:
<span class="math-container">$$
F_{X_2}(a) = \Pr(X_2 \le a) = \Pr(X_1 \le a \text{ and } X_2 \le a) = \Pr(X_1 \le a)\Pr(X_2 \le a) = a^2.
$$</span>
This assumes that the two uniformly random variables are independent.</p>
<p>Then the density is given by:
<span class="math-container">$$
f_{X_2}(a) = \frac{\partial F_{X_2}(a)}{\partial a} = 2 a.
$$</span></p>
<p>The expected value of <span class="math-container">$X_2^{\frac{m}{1-m}}$</span> is then:
<span class="math-container">$$
\int_0^1 a^{\frac{m}{1-m}} f_{X_2}(a) da = \int_0^1 a^{\frac{m}{1-m}} 2 a\, da,\\
=\int_0^1 2 a^{\left(\frac{m}{1-m} + 1\right)} da = \int_0^1 2 a^{\frac{1}{1-m}} da = 2 \frac{1-m}{2-m}
$$</span></p>

CC BY-SA 4.0
2021-04-29T12:11:12.790Z
This looks perfectly great! I just wanted to ask one last thing. When calculating the pdf, do we always take the original form? Like here X is U(0,1). But we need to calculate ${X_n} ^{m/(1-m)}$. So, even if X is raised to power something, we calculated pdf of X and not X raised to the power, is this correct
Elina Gilbert
CC BY-SA 4.0
2021-04-29T12:18:12.183Z
Yes. If $X$ is a rv then the expected value of $g(X)$ is given by $\int g(x) f_X(x) dx$ where $f_X$ is the pdf of $X$. In your case, $g(x) = x^{m/(1-m)}$.
tdm
CC BY-SA 4.0
2021-04-29T12:18:13.230Z
And sorry, I forgot to mention that m is between 0 to 1. I edited the question. Your assumption, m<1 is correct.
Elina Gilbert
CC BY-SA 4.0
2021-04-29T12:22:17.540Z
I can't thank you enough! You cleared all my doubts
Elina Gilbert
CC BY-SA 4.0
2021-04-29T12:03:03.737Z
2021-04-29T12:03:03.737Z
tdm
<p>I have <span class="math-container">$X$</span> ~ <span class="math-container">$U(0,1)$</span> interval.</p>
<p>Let n=2, i.e. <span class="math-container">$X_1 &lt; X_2$</span></p>
<p>I have to calculate the expected value of <span class="math-container">${X_2}^{m/(1-m)}$</span>.</p>
<p>Where, <span class="math-container">$0≤m≤1$</span></p>
<p>I want to confirm if I have calculated it correctly?</p>
<p><span class="math-container">$$\int_0^1 2x^{2m/(1-m)} \,dx = \frac{2(1-m)}{(1+m)}$$</span></p>
<p>Is the calculation correct?</p>
<p>Thanks!</p>

<statistics><probability><expectations>
Expected value of order statistics for uniform distribution
Elina Gilbert
# Expected value of order statistics for uniform distribution

(Asked by: username_0 on 2021-04-29)

I have $X$ ~ $U(0,1)$ interval.

Let n=2, i.e. $X\_1 < X\_2$

I have to calculate the expected value of ${X\_2}^{m/(1-m)}$.

Where, $0≤m≤1$

I want to confirm if I have calculated it correctly?

$$\int\_0^1 2x^{2m/(1-m)} \,dx = \frac{2(1-m)}{(1+m)}$$

Is the calculation correct?

Thanks!

**Tags:** statistics, probability, expectations


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## 1 answers

### Answer by username_1 on 2021-04-29. Score: 1 | Accepted answer

I guess $m < 1$. Otherwise $x^{1/(m-1)} \to \infty$ as $x \to 0$.

There's a possibility that I made a mistake, so let me know if you spot one.

Let's first compute the cumulative distribution function of $X\_2$:
$$
F\_{X\_2}(a) = \Pr(X\_2 \le a) = \Pr(X\_1 \le a \text{ and } X\_2 \le a) = \Pr(X\_1 \le a)\Pr(X\_2 \le a) = a^2.
$$
This assumes that the two uniformly random variables are independent.

Then the density is given by:
$$
f\_{X\_2}(a) = \frac{\partial F\_{X\_2}(a)}{\partial a} = 2 a.
$$

The expected value of $X\_2^{\frac{m}{1-m}}$ is then:
$$
\int\_0^1 a^{\frac{m}{1-m}} f\_{X\_2}(a) da = \int\_0^1 a^{\frac{m}{1-m}} 2 a\, da,\\
=\int\_0^1 2 a^{\left(\frac{m}{1-m} + 1\right)} da = \int\_0^1 2 a^{\frac{1}{1-m}} da = 2 \frac{1-m}{2-m}
$$

> **Comments:**
> **username_0**: This looks perfectly great! I just wanted to ask one last thing. When calculating the pdf, do we always take the original form? Like here X is U(0,1). But we need to calculate ${X_n} ^{m/(1-m)}$. So, even if X is raised to power something, we calculated pdf of X and not X raised to the power, is this correct  
> **username_1**: Yes. If $X$ is a rv then the expected value of $g(X)$ is given by $\int g(x) f_X(x) dx$ where $f_X$ is the pdf of $X$. In your case, $g(x) = x^{m/(1-m)}$.  
> **username_0**: And sorry, I forgot to mention that m is between 0 to 1. I edited the question. Your assumption, m<1 is correct.  
> **username_0**: I can't thank you enough! You cleared all my doubts  

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