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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!

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1 Answer 1

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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: $$ \int_0^1 a^{\frac{m}{1-m}} f_{X_2}(a) da = \int_0^1 a^{\frac{m}{1-m}} 2 a\, da,\\ =\int_0^1 2 a^{\left(\frac{m}{1-m} + 1\right)} da = \int_0^1 2 a^{\frac{1}{1-m}} da = 2 \frac{1-m}{2-m} $$

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  • $\begingroup$ 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 $\endgroup$ Commented Apr 29, 2021 at 12:11
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    $\begingroup$ 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, $g(x) = x^{m/(1-m)}$. $\endgroup$
    – tdm
    Commented Apr 29, 2021 at 12:18
  • $\begingroup$ And sorry, I forgot to mention that m is between 0 to 1. I edited the question. Your assumption, m<1 is correct. $\endgroup$ Commented Apr 29, 2021 at 12:18
  • $\begingroup$ I can't thank you enough! You cleared all my doubts $\endgroup$ Commented Apr 29, 2021 at 12:22

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