Question

Consider the following probability density function: -x-1/2e-z/2 for x > 0. f(x) = the area under the curve (integral) is equal to one, then: i) Compute the mean of the function numerically based on the principle: rf (x) dr ES Where S is the set of values on which the function is defined i Compute the median y where: f(z) dz = Where m is the minimum value on which the function is defined.

Answer 1

Answer:)

We have to first prove that the area under the curve is 1. Then,

A = $\int_{0}^{\infty} \dfrac{1}{\sqrt{2\pi}} x^{-1/2} e^{-x/2} dx \xrightarrow{x = 2z} \dfrac{1}{\sqrt{\pi}} \cdot \int_{0}^{\infty} z^{-1/2} e^{-z} dz = \dfrac{1}{\sqrt{\pi}} \cdot \Gamma(1/2) = 1$

where the gamma function is defined as $\Gamma(z) = \int_0^\infty x^{z-1} e^{-x}\, dx$

and it is known that $\Gamma(1/2) = \int_0^\infty x^{-1/2} e^{-x}\, dx = \sqrt{\pi}$

Answer to Part 1.):

Now, we get that:

2 Mean = 1 0 2 1

We have used the fact that the Gamma function has the property that $\Gamma(n+1) = n \Gamma(n)$

Answer to Part 2.)

Now, we get that:

1/2/2 0 0

There is no simple closed for in order for us to find $\gamma$, and we must use numerical methods in order to estimate $\gamma$. On MATLAB, this answer is given by gaminv(0.5,0.5,2) since in our case we need to find the inverse to the gamma cumulative distribution function with parameters 0.5 and 2 with a given probability of 0.5, and the answer is

γ 0.454936423119573

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1/2 e-1/2 dl

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1/2/2 0 0

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3a0.8 (a =-) 3a + 0.2, 0.150639209672741

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1/2/2 0 0

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