# Let $X_1, X_2, \dots,X_n$ independent random variables.

Let $$X_1, X_2, \dots,X_n$$ independent random variables such that for $$1\le i \le n$$ the distribution of $$X_i$$ is $$\mathbb{P}(X_i=2^i)=\frac{1}{2^i}, \space \space \mathbb{P}(X_i=1)=\frac{1}{2}-\frac{1}{2^{i+1}}, \space \space\mathbb{P}(X_i=-1)=\frac{1}{2}-\frac{1}{2^{i+1}}$$

For $$1\le i \le n$$, we'll define the event $$A_i = \{X_i\ne2^i\}$$ and the random variable $$Y_i=X_i\cdot \mathbb{1}_{A_i}$$.

(a) Prove: For all $$i$$: $$\mathbb{E}Y_i=0$$

(b) For $$1\le k \le n$$, we'll define the event $$B_k$$ that there exists $$k\le i \le n$$, such that: $$X_i\ne Y_i$$. Prove the following inequality: $$\mathbb{P}(B_k)\le\frac{1}{2^{k-1}}$$

My try:

(a) $$Y_i$$ takes the values $$0, 1$$ or $$-1$$:

$$\mathbb{E}Y_i=\mathbb{P}(Y_i=1)\cdot1+\mathbb{P}(Y_i=-1)\cdot(-1)+\mathbb{P}(Y_i=0)\cdot0$$

Since: $$\mathbb{P}(Y_i=1)=\mathbb{P}(X_i=1)\cdot\mathbb{P}(\mathbb{1}_{A_i}=1)=\mathbb{P}(X_i=-1)\cdot\mathbb{P}(\mathbb{1}_{A_i}=1)=\mathbb{P}(Y_i=-1)$$

Then, the result above is 0. Can anyone validate that proof?

(b) I'm struggling a bit with the definitions. How is that possible that $$X_i=Y_i$$ as $$Y_i$$ takes the values: $$0, 1$$ or $$-1$$ and $$X_i$$ takes $$2^i, 1$$ or $$-1$$, so they would never be equal.

I hope you could help me get the definitions straight.

• Why the downvote? Please let me know how can I improve my question. Commented May 5, 2018 at 10:17
• b) does not talk about $X_i$ and $Y_i$ being equal with probability 1. The event $\{X_i=Y_i\}$ definitely makes sense, right? Commented May 5, 2018 at 12:04

Observe that $$B_k=\bigcup_{i=k}^nA_i^c$$ because the only way that $$X_k$$ and $$Y_k$$ differ is that the indicator of $$A_i$$ vanishes. Since the probability of $$A_i^c$$ is $$2^{-i}$$, it follows that $$\Pr\left(B_k\right)\leqslant \sum_{i=k}^n2^{-i}=2^{-k}\underbrace{\sum_{j=0}^{n-k}2^{-j}}_{\leqslant 1/2}.$$