MHB Compute Expectation for $X_{1}^\frac{1}{2}$ with Family $f(x,\theta)$

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To compute the expectation \( E(X_{1}^{1/2}) \) for the given density family \( f(x,\theta) = \frac{exp(-\sqrt{x}/\theta)}{2\theta^2} \), the integral is transformed into a more manageable form. The expectation can be expressed as \( E(X_1^{1/2}) = \int_0^{\infty} \sqrt{x} \frac{exp(-\sqrt{x}/\theta)}{2\theta^2} dx \). By substituting \( t = \sqrt{x} \), the integral simplifies to \( \int_0^{\infty} t^2 \frac{exp(-t/\theta)}{\theta^2} dt \). This transformation allows for easier computation of the expectation, addressing the initial concern about integration by parts. The discussion highlights the importance of substitution in evaluating complex integrals.
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consider a density family $f(x,\theta)=\frac{exp(-{\sqrt{x}}/{\theta})}{2{\theta}^2}$.

Let $X_{1}$ have the density above. Compute $E(X_{1}^\frac{1}{2})$.

Integration by parts doesn't work since the derivative of ${\sqrt{x}}$ never vanishes, so how do I compute the expectation?
 
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Have you got this?

$$E(X_1^{1/2})=\int_0^{\infty} \sqrt{x}\exp(-\sqrt{x}/\theta)/2\theta^2dx=\int_0^{\infty} t^2\exp(-t/\theta)/\theta^2dt$$
 
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Thanks, I'm a bit out of practice with integrals
 
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