$Z_1\overset{P}{\to}\infty \implies h(Z_1, Z_2)\overset{P}{\to}\infty$ for $Z_1, Z_2$ independent

Question

Consider two independent real random variables $Z_1, Z_2$ and a real function $h:\mathbb{R}^2\to\mathbb{R}$ such that $\lim_{u\to\infty}h(u, z_2) = \infty$ for any $z_2\in\mathbb{R}$. Does it hold that for any $c\in\mathbb{R}$ is $$ \lim_{u\to\infty} P(h(Z_1, Z_2)\leq c\mid Z_1>u)=0? $$

I feel like an approach using nested conditional expectation should be helpful, but I can not formalize it properly. Also, should $h$ be continuous for this to be valid?

Answer 1

Yes (no need for continuity of $h$). I will assume that $\mathbb P(Z_1>u)>0$ for all $u\in\mathbb R$, so that the conditional probability is well defined.

Let $c\in\mathbb R$. Denoting by $\mu$ the distribution of $Z_2$, we have that $$ \begin{align*} \mathbb P(h(Z_1,Z_2)\le c\vert Z_1>u)&=\frac{\mathbb P(h(Z_1,Z_2)\le c,Z_1>u)}{\mathbb P(Z_1>u)}\\ &=\int_{\mathbb R}\frac{\mathbb P(h(Z_1,z_2)\le c,Z_1>u)}{\mathbb P(Z_1>u)}\,\mu(dz_2). \end{align*} $$

Let $z_2\in\mathbb R$. As $h(u,z_2)$ goes to $+\infty$ with $u$, the integrand is $0$ for $u$ large enough. We deduce that the integrand converges pointwise to $0$ as $u\to+\infty$. As it is bounded by $1$, the dominated convergence theorem yields that the integral converges to $0$ as $u\to+\infty$.

Answer 2

Let $c\in \mathbb R$. First, note that by assumption on $h$, for any $z_2\in\mathbb R$ there exists $u_0(=u_0(z_2))\in\mathbb R$ such that $h(u,z_2)>c $ for all $u \ge u_0$. This is an important fact that we'll need later.

Second, observe that by the law of total expectation, we have the identity :

$$\mathbb P(h(Z_1, Z_2)\leq c\mid Z_1>u) = \mathbb E\left[ \mathbb P(h(Z_1, Z_2)\leq c\mid Z_1>u, Z_2)\right] \tag1$$

In the above, I condition on both the event $\{Z_1>u\} $ and on the random variable $Z_2$.

Now, given $Z_2$, we see from what we've said at the beginning of this post that there exists $u_0(Z_2)\in\mathbb R$ such that $u\ge u_0 \implies h(Z_1,Z_2) > c$ whenever $Z_1 > u$. In other words, given $Z_2$, the set $E(c,u) := \{h(Z_1, Z_2)\le c\text{ AND } Z_1>u \} $ is empty for all sufficiently large $u$.

Now, developing the conditional expectation as in this answer, we get :

$$\begin{align}\mathbb P(h(Z_1, Z_2)\leq c\mid Z_1>u) &= \mathbb E\left[ \mathbb P(h(Z_1, Z_2)\leq c\mid Z_1>u, Z_2)\right]\\ &= \mathbb E\left[ \mathbf 1\{h(Z_1, Z_2)\leq c\}\mid Z_1>u, Z_2\right]\\ &=\frac{\mathbb E\left[ \mathbf 1\{h(Z_1, Z_2)\leq c \text{ AND } Z_1 > u\}\mid Z_2\right]}{\mathbb E\left[ \mathbf 1\{ Z_1 > u\}\mid Z_2\right]}\\ &=\frac{\mathbb E\left[ \mathbf 1\{E(c,u)\}\mid Z_2\right]}{\mathbb E\left[ \mathbf 1\{ Z_1 > u\}\mid Z_2\right]}\label2 \tag2\end{align} $$

EDIT : I might be wrong, but it seems that I can't apply the dominated convergence theorem with this approach because of the denominator in $\eqref 2$. For my argument to go through, I instead need to assume that $u^* := \sup \{u_0 (z_2)\mid z_2\in \mathbb R\}<\infty$, then we get that $$\begin{align*} \lim_{u\to\infty} \mathbb P(h(Z_1, Z_2)\leq c\mid Z_1>u) &=\lim_{u\to\infty\\ u>u^*} \mathbb P(h(Z_1, Z_2)\leq c\mid Z_1>u)\\ &=\lim_{u\to\infty\\ u>u^*} \frac{\mathbb E\left[ \mathbf 1\{E(c,u)\}\mid Z_2\right]}{\mathbb E\left[ \mathbf 1\{ Z_1 > u\}\mid Z_2\right]}\\ &=\lim_{u\to\infty\\ u>u^*}\frac{0}{\mathbb E\left[ \mathbf 1\{ Z_1 > u\}\mid Z_2\right]}\\ &=0 \end{align*} $$

Which is the desired conclusion, up to that extra assumption on $h$ of course.

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Remark : in equation $\eqref 2$, we have $\mathbb E\left[ \mathbf 1\{ Z_1 > u\}\mid Z_2\right] \equiv \mathbb P (Z_1 > u \mid Z_2) $ in the denominator, and that probability may vanish depending on the support of $Z_1$. We thus need to set $\mathbb P(A\mid B)\equiv 0$ whenever $\mathbb P(B\mid Z_2)=0$ AND $\mathbb P(A\mid Z_2) = 0$ in order to avoid division by zero (and ensuring that the desired result still holds).