I am trying to prove the following:

Let $X_{n}$ be a sequence of random variables converging in probability to some random variable $X$. Furthermore $P(|Xn|>k)=0$ for all n and some $k>0$.

Let $Y_{n}$ be a sequence in $L^{1}(\Omega)$. Assume that there exists a real number $\lambda$ such that $E(Y_{n}) = \lambda$ for all n and $\sup |Y_{n}| \le \eta$ for some $ \eta \in L^{1}(\Omega) $.

Prove that if $X=c$ then $lim_{n \to \infty} E(Y_{n}X_{n})=c \lambda$.

How do I show this if the sequences are not independent?

  • 1
    $\begingroup$ Are there any other assumptions about $X_n$'s (e.g. bounded or u.i.)? Even if $Y_n$'s are independent of $X_n$'s, there is no guarantee that $\mathbb{E}[X_n]\to c$. $\endgroup$ – d.k.o. Nov 7 '16 at 4:14
  • $\begingroup$ There was a first part to the problem which states that $P(|X_{n}|>k)=0$ for all n and some $k > 0.$ (The first part asked to prove that for all $r >0$$ lim_{n \to \infty} X_{n} = X$ in $L^{r}(\Omega)$.) $\endgroup$ – user75514 Nov 7 '16 at 19:32
  • 2
    $\begingroup$ Just edit the question! $\endgroup$ – d.k.o. Nov 7 '16 at 20:04

Here's something to get you started. $$\begin{align}|E(Y_nX_n) - c \lambda| &= |E(Y_nX_n) - c E(Y_n)| = |E(Y_n(X_n - c))| \\ &\leq |E(Y_n(X_n - c)\mathbf{1}_{|X_n - c| \geq \epsilon})| + |E(Y_n(X_n - c)\mathbf{1}_{|X_n - c| < \epsilon})| \\ &\leq E(|\eta| |X_n - c|\mathbf{1}_{|X_n - c| \geq \epsilon}) + \epsilon |\lambda| \end{align}$$

  • $\begingroup$ So we are going to show that $|E(Y_{n}X_{n}) -c \lambda| \to 0$ as $n \to \infty$, right? Let me make sure I can fill in the piece of your hint. By your last inequality, we know that $X_{n} \to c$ as $n \to \infty$ with probability 1 since $X_{n}$ converges to $X$ in probability. Then if we take the limit as $n \to \infty$ on the last inequality and choose $\epsilon$ to be 0, we have the first term does not blow up since $\eta \in L^{1}$ and tends to 0 since $X_{n}$ converges to $c$ in probability and the second term is zero by choosing epsilon arbitrarily small. $\endgroup$ – user75514 Nov 7 '16 at 19:53


$$ \mathbb{E}[Y_nX_n]=\mathbb{E}[Y_n(X_n-c)]+c\lambda. $$

However, since $\{X_n\}$ is uniformly bounded (by your comment), the DCT implies that

$$ |\mathbb{E}[Y_n(X_n-c)]|\le \mathbb{E}\eta|X_n-c|\to 0 \quad\text{as } n\to \infty. $$

  • $\begingroup$ How do we know that $E(Y_{n}X_{n})=E(Y_{n}(X_{n} -c)) +c \lambda$? $\endgroup$ – user75514 Nov 7 '16 at 20:58
  • $\begingroup$ @user75514 $$ \mathbb{E}[Y_nX_n]=\mathbb{E}[Y_n(X_n-c+c)]=\mathbb{E}[Y_n(X_n-c)]+\mathbb{E}[Y_nc] $$ $\endgroup$ – d.k.o. Nov 7 '16 at 21:17

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