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My problem is that I have the distribution

$f_{z}(x)=\dfrac{2z^2}{x^3}$, $0<z<x$ and I have to prove that $T(X_1,\ldots,X_n\mid z)= \dfrac{1}{z}\min(X_1,\ldots,X_n)$ is a pivotal quantity.

I have calculated the distribution of $\min(X_1,\ldots,X_n)$ and my result is $\dfrac{z^{2n}}{x^{2n+1}}n$ so I dont get the result i have been asked.

¿I have calculate the distribution wrong? thanks

Michael Hardy
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Gabrielle
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  • Please show your working. How did you obtain that distribution? – dreamer Jun 06 '13 at 17:57
  • ok, $f_{min}(x)=(1-F(x))^{n-1}nf_{z}(x)$and my $F(x)=1-\dfrac{z^{2}}{x^2}$making the integral between z and x – Gabrielle Jun 06 '13 at 17:58
  • ok thanks, but it isnt enough to divide by z and you have the final distribution of the pivotal??? – Gabrielle Jun 06 '13 at 18:04
  • In general not. If $Y=u(X)$ defines a one-to-one transformation then $f_Y(y)=f_X(w(y))|{\dfrac{d}{dy}w(y)}|$, where $w(y)$ denotes the inverse function of $u(X)$. – dreamer Jun 06 '13 at 18:07

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For $z\le a \le b$ you have $$ \Pr(a\le X_1\le b) = \int_a^b \frac{2z^2}{x^3} \, dx = \left.\frac{-z^2}{x^2}\right|_{w=a}^{w=b} = \frac{z^2}{a^2} - \frac{z^2}{b^2}. $$ As $b\to\infty$ and $a\to0$, this approaches $1$, so you do have a probability density on $(z,\infty)$.

From this it follows that $$ \Pr(X_1 > x) = \frac{z^2}{x^2}\text{ for }x\ge z. $$ $$ \Pr(\min > x) = \Pr(X_1,\ldots,X_n \text{ all } > x) = \left( \Pr(X_1> x) \right)^n = \left(\frac{z^2}{x^2}\right)^n = \frac{z^{2n}}{x^{2n}}\text{ for }x\ge z. $$ Now we want $$ \Pr\left(\frac1z \min > x\right) \text{ for }x\ge 1. $$ We have $$ \Pr\left(\frac1z \min > x\right) = \Pr(\min > xz) = \frac{z^{2n}}{(xz)^{2n}} = \frac{1}{x^{2n}}. $$ That does not depend on $z$.

Michael Hardy
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  • Just out of curiousity, why do you prefer to write the distributions in non-standard form (I mean, in the form $P(\bullet > a)$, and not as a CDF or PDF)? – dreamer Jun 06 '13 at 18:49
  • Because in this case the probability of being greater than something is what was needed. – Michael Hardy Jun 06 '13 at 18:51
  • Why? Isn't the only thing that needs to be shown that the distribution is not dependent on parameter $z$? Wouldn't that also be possible using the PDF or CDF? – dreamer Jun 06 '13 at 18:54
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    The probability that the min is greater than something is merely the product of the probabilities that the separate observations are greater than that thing. If I'd said "less than" instead of "greater than", then it wouldn't be just multiplying: I'd have to subtract from $1$, then multiply, then subtract from $1$ again. Just extra complications that aren't enlightening. I believe in making things look only as complicated as they really are. – Michael Hardy Jun 06 '13 at 18:56
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    Ok, I agree, you have a point there :). – dreamer Jun 06 '13 at 18:58