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I am currently preparing for an upcoming Discrete choice models exam. I have come across an exercise where I need to calculate the conditional expected value.

Given:

$X$~$Pareto (a, b)$. Find $E(X | X > 11)$ when $(a, b) = (\frac{11}{2}, \frac{5}{2})$.

What I have so far:

$PDF = \frac{ab^a}{u^{a+1}}, \hspace{2mm} P(X > 11) = (\frac{a}{11})^b$

Also, $\hspace{2mm} E(X) = \frac{ab}{a-1}$

Problem: I do not quite understand what is the process for calculating a conditional expected value.

RokasR
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  • Can you find the conditional density of $X$ given $X>11$? – StubbornAtom May 03 '23 at 10:26
  • Find the expectation of $X1_{X>11}$ and apply:$$\mathbb E[X\mid X>11]=\frac{\mathbb EX1_{X>11}}{P(X>11)}$$ – drhab May 03 '23 at 10:51
  • Regarding the answer from @drhab, I get that:

    $E(X|X>11) = \frac{E(X) \cdot 1_{X>11}}{P(X>11)} = \frac{ab}{a-1} \cdot (\frac{11}{a})^b = \frac{ab \cdot 11^b}{(a-1)a^b}$

    Given that $(a, b) = (\frac{5}{2}, \frac{11}{2})$, I get:

    $E(X|X>11) = 31710.36$

    Did I understand the procedure correctly?

     – RokasR May 03 '23 at 11:45

2 Answers2

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The calculation of the probability $\Pr[X > 11]$ is not consistent with your definition of the PDF. If you define $$f_X(x) = \frac{ab^a}{x^{a+1}}, \quad x \ge b,$$ then $$\Pr[X > x] = \int_{t = x}^\infty f_X(t) \, dt = \int_{t = x}^\infty \frac{ab^a}{t^{a+1}} \, dt = ab^a \left[ -\frac{1}{at^a} \right]_{t=x}^\infty = ab^a \left( 0 + \frac{1}{ax^a} \right) = \left(\frac{b}{x}\right)^a,$$ and not $(a/x)^b$ as you stated. So you need to make sure that you are not mixing up the parameters.

That said, we can compute the conditional expectation as follows:

$$\operatorname{E}[X \mid X > x] = \frac{\int_{t=x}^\infty t f_X(t) \, dt}{\Pr[X > x]} = \left(\frac{x}{b}\right)^a \int_{t=x}^\infty \frac{ab^a}{t^a} \, dt.$$ The remainder of the computation is left to you as an exercise; then setting $x = 11$, $a = 5/2$, $b = 11/2$ gives the desired numerical answer.

heropup
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  • Thanks for the correction. I used $F_X(u) = 1 - (\frac{a}{u})^b$ as a PDF in order to calculate $P(X>11)$.

    I switched to $F_X(u) = \frac{ab^a}{u^{a+1}}$. Now, I get that $P(X>11) = P(X \leq 11) = 1 - P(X \leq 11) = 1 - \frac{ab^a}{11^{a+1}}$.

     – RokasR May 03 '23 at 15:25
  • Also, I get that $\int_{11}^{\infty} u \cdot F_X(u) ,du = \frac{ab^a}{a-1} \cdot 11^{-a+1}$. – RokasR May 03 '23 at 15:43
  • I used a different method on both calculating $P(X > 11)$ as well as the integral. Are my calculations correct? – RokasR May 03 '23 at 15:54
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Let $X\sim\mathsf{Pareto}(a,b)$ and let $c>b$.

Then for $x>b$ we have $P(X>x)=\left(\frac{b}{x}\right)^a$ so for $x>c$ we find:$$P(X>x|X>c)=\frac{P(X>x)}{P(X>c)}=\left(\frac{c}{x}\right)^a$$showing that the conditional distribution is again Pareto but this time with parameters $a$ and $c$.

So if $a>1$ then we conclude that:$$\mathbb E[X|X>c]=\frac{ac}{a-1}$$

Also you can go for $$\mathbb E[X|X>c]$$ directly according to the formula:$$\mathbb E[X|X>c]=\frac{\mathbb EX1_{X>c}}{P(X>c)}$$ Here the numerator is $\frac{ac}{a-1}\left(\frac{b}{c}\right)^a$ and the denominator is $\left(\frac{b}{c}\right)^a$.

drhab
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  • I don't think that I understand the concept. In my calculations, I use $PDF: f_X(u) = \frac{ab^a}{u^{a+1}}$ to get $EX \cdot 1_{X>11}$ and $CDF: F_X(u) = 1 - (\frac{a}{u})^b$ to get $P(X > 11)$. Is that right? – RokasR May 03 '23 at 19:27
  • Yes (as I state in the second part of my answer). Do your answers for $\mathbb EX1_{X>c}$ and $P(X>c)$ agree with mine if $c=11$? – drhab May 04 '23 at 06:51
  • In my calculations, I get that $P(X>11) = (\frac{a}{c})^b = (\frac{5.5}{11})^{2.5} = 0.18$ while $(\frac{b}{c})^a = (\frac{2.5}{11})^{5.5} = 2.89 \cdot 10^{-4}$.

    The same goes for numerator, the answer in my calculations is different.

     – RokasR May 05 '23 at 09:47
  • Using the PDF you mention we arrive at:$$P\left(X>c\right)=\int_{c}^{\infty}\frac{ab^{a}}{u^{a+1}}du=ab^{a}\int_{c}^{\infty}u^{-a-1}du=ab^{a}\left[\frac{u^{-a}}{-a}\right]_{c}^{\infty}=ab^{a}\left(0-\frac{c^{-a}}{-a}\right)=\left(\frac{b}{c}\right)^{a}$$In the other answer you are already warned not to switch parameters. – drhab May 06 '23 at 08:54