Conditional expectation $E\left[X|X\geq aY\right]$ if $X$ and $Y$ are independent Fréchet

Question

Let $X$, $Y$ be two independent continuous random variables with support on $\mathbb{R}_+$. More precisely, assume that $X$ and $Y$ are distributed Fréchet with location parameter $0$, shape parameter $\alpha$ and scale parameters $s_X$ and $s_Y$ respectively (hence the only difference between these distributions is the scale parameter). Let $a$ denote a positive scalar. How to compute the conditional expectation $$ E\left[X|X\geq aY\right]\ ? $$

My first approach was to assume a fixed $Y$ and then integrate with respect to $Y$, i.e.

$$ E\left[X|X\geq aY\right] =\int_{0}^{\infty}E\left[X|X\geq ay\right]f_{y}\left(y\right)dy =\int_{0}^{\infty}\left[\int_{ay}^{\infty}xf_{x}\left(x\right)dx\right]f_{y}\left(y\right)dy $$

But I got stuck with those huge integrals after substituting $f_x$ and $f_y$ for the Fréchet distribution.

I also tried finding the conditional distribution of $X$ for $X\geq aY$, i.e. $f_{X|X\geq aY}(x,y)$, so I can write

$$ E\left[X|X\geq aY\right] = \int_{0}^\infty x f_{X|X\geq aY}\left(x|y\right)dx $$

But I don't know how to obtain the joint distribution given that the condition depends on both $X$ and $Y$.

Answer 1

The joint distribution is not necessary, rather use the fact that, if $X$ and $Y$ are independent Fréchet $(a,s_X,0)$ and $(a,s_Y,0)$ respectively, then $$(X,Y)=(s_XU^{-1/a},s_YV^{-1/a})$$ where $(U,V)$ are i.i.d. standard exponential. Thus, for every positive $t$, $$[X>tY]=[V>rU]\qquad r=(ts_Y/s_X)^a$$ and, by independence, $$P(V>rU\mid U)=e^{-rU}$$ hence $$E(X\mid X>tY)=s_XE(U^{-1/a}\mid V>rU)=s_X\frac{E(U^{-1/a}\mathbf 1_{V>rU})}{P(V>rU)}=s_X\frac{E(U^{-1/a}e^{-rU})}{E(e^{-rU})}$$ Now, the numerator and the denominator are simple integrals, which evaluate to $$E(U^{-1/a}e^{-rU})=(r+1)^{1/a-1}\Gamma(1-1/a)\qquad E(e^{-rU})=(r+1)^{-1}$$ for every $a>1$ (otherwise $X$ is not integrable hence the expectation in the numerator is infinite). Finally, for every $a>1$, $$E(X\mid X>tY)=s_X(r+1)^{1/a}\Gamma(1-1/a)$$

Answer 2

$$E(X | X > \alpha Y) = \frac{E(X1_{X>\alpha Y})}{P(X>\alpha Y)}$$

$$E(X1_{X>\alpha Y}) = E(E(X1_{X>\alpha Y}|X)) = E\left(XF_{Y}\left(\frac{X}{\alpha}\right)\right)$$

$$\begin{align} E\left(XF_{Y}\left(\frac{X}{\alpha}\right)\right) &= \int_{0}^{\infty} xe^{-(x/(s_y\alpha))^{-\alpha}}\frac{\alpha}{s_x}\left(\frac{x}{s_x}\right)^{-1-\alpha}e^{-(x/s_x)^{-\alpha}} dx \\\\ &= \frac{\alpha}{s_x^{-\alpha}}\int_{0}^{\infty} x \cdot x^{-1-\alpha}e^{-\left(\frac{x}{\frac{s_xs_y\alpha}{(s_x^{-\alpha} + (s_y\alpha)^{-\alpha})^{-1/\alpha}}}\right)^{-\alpha}} dx \\\\ &= \frac{\alpha}{s_x^{-\alpha}}\cdot \frac{S^{-\alpha}}{\alpha}\cdot S\cdot \Gamma\left(1-\frac{1}{\alpha}\right) \\\\ &= \frac{S^{1-\alpha}}{s_x^{-\alpha}}\cdot \Gamma\left(1-\frac{1}{\alpha}\right) \text{ if } \alpha > 1\end{align}$$

where,

$$S = \frac{s_xs_y\alpha}{(s_x^{-\alpha} + (s_y\alpha)^{-\alpha})^{-1/\alpha}} = (s_x^\alpha + (\alpha s_y)^\alpha)^{\frac{1}{\alpha}}$$

And,

$$\begin{align} P(X > \alpha Y) &= \int_{0}^\infty \int_{0}^{\frac{x}{\alpha}} f_{X}(x)f_{Y}(y)dxdy \\\\ &= \int_{0}^\infty f_{X}(x)F_{Y}\left(\frac{x}{\alpha}\right)dx \\\\ &= \int_{0}^{\infty} \frac{\alpha}{s_x}\left(\frac{x}{s_x}\right)^{-1-\alpha}e^{-(x/s_x)^{-\alpha}} e^{-(x/(s_y\alpha))^{-\alpha}} dx \\\\ &= \frac{\alpha}{s_x^{-\alpha}}\int_{0}^{\infty} x^{-1-\alpha}e^{-\left(\frac{x}{\frac{s_xs_y\alpha}{(s_x^{-\alpha} + (s_y\alpha)^{-\alpha})^{-1/\alpha}}}\right)^{-\alpha}} dx \\\\ &= \frac{\alpha}{s_x^{-\alpha}}\cdot \frac{S^{-\alpha}}{\alpha} = \frac{S^{-\alpha}}{s_x^{-\alpha}} \end{align}$$

Finally,

$$E(X|X>\alpha Y) = S \cdot \Gamma\left(1-\frac{1}{\alpha}\right) \text{ if } \alpha > 1$$