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Take the following random variables $Y,\epsilon_0,\epsilon_1,V_0,V_1$, with finite supports $\mathcal{Y}\equiv \{0,1\},\mathcal{E}_0, \mathcal{E}_1,\mathcal{V}_0,\mathcal{V}_1$, respectively.

Suppose $$P_Y(1|e_0,e_1,v_0,v_1)=\mathbb{1}\{e_0+v_0\leq e_1+v_1\}$$ $\forall e_0\in \mathcal{E}_0$,$\forall e_1\in \mathcal{E}_1$,$\forall v_0\in \mathcal{V}_0$,$\forall v_1\in \mathcal{V}_1$, where $\mathbb{1}\{a\leq 0\}$ is $1$ if $a\leq 0$ and $0$ otherwise.

I want to derive the unconditional probability distribution of $Y$.

I can use the law of total probability, that is $$ (\diamond) \text{ }\text{ }\text{ }P_Y(1)=\sum_{e_0,e_1,v_1,v_0} \mathbb{1}\{e_0+v_0\leq e_1+v_1\} \times P_{\epsilon_0,\epsilon_1,V_0,V_1} $$

Question: Let $W_0\equiv \epsilon_0+V_0$ and $W_1\equiv \epsilon_1+V_1$. My intuition without going through the law of total probability is that $$ (\diamond \diamond) \text{ }\text{ }\text{ }P_Y(1)= \sum_{w_0,w_1} \mathbb{1}\{w_0\leq w_1\}\times P_{W_0, W_1}(w_0, w_1) $$ In other words, knowing the joint distribution of the sums, $P_{W_0, W_1}$ (rather than the joint distribution of each term, $P_{\epsilon_0,\epsilon_1,V_0,V_1}$) is sufficient. Is this correct? If yes, how do we show that $(\diamond)=(\diamond \diamond)$?

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Yes, it is sufficient. Letting $w_0=e_0+v_0$ and $w_1=e_1+v_1$, you can safely add functions of your RVs to the given side of the probability formulas, and then rename them: $$\begin{align}P_Y(1|e_0,e_1,v_0,v_1)&=P_Y(1|e_0,e_1,v_0,v_1,e_0+v_0,e_1+v_1)\\&=P_Y(1|e_0,v_0,e_1,v_1,w_0,w_1)\\&=\mathbb{1}\{w_0\leq w_1\}\end{align}$$ Since, the RHS probability doesn't depend further on $e_i,v_i$, we can safely omit them in LHS: $$P_Y(1|w_0,w_1)=\mathbb{1}\{w_0\leq w_1\}$$ Now, you can rewrite your second formula, i.e. $\diamond\diamond$, as follows by changing the first multiplicand inside the summation expression: $$P_Y(1)=\sum_{w_0,w_1}P_Y(1|w_0,w_1)\times P_{W_0,W_1}(w_0,w_1)$$ this is always true due to Total Probability Law, which also applies to $(\diamond)$ saying also that these two are equivalent because they are both different versions of $P_Y(1)$.

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