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The wording of my question may be unclear, since I did not know how to else to post it, but any help would be greatly appreciated, and feel free to edit it!

I am really confused about how to approach the following question from Sheldon Ross' A first course in probability, chapter 4.

The question is:

Let $N$ be a non-negative integer-valued random variable. For non-negative values, $a_{j}, j\ge 1$, show that

$$\sum_{j=1}^{\infty} (a_{1} + a_{2} + ...... a_{j})(P(N=j)) = \sum_{j=1}^{\infty} a_{i}P(N\ge i)$$

I get this part, what I don't get is the following:

$$E(N(N+1)) = 2\sum_{j=1}^{\infty}iP(N\ge i)$$

User203940
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1 Answers1

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We can rewrite what you have as

$$ \sum_{j=1}^\infty \sum_{k=1}^j a_k P(X=j).$$ Notice that

$$ \sum_{j=1}^\infty \sum_{k=1}^j a_k P(X=j) = a_1 P(X=1) + a_1P(X=2) + a_2 P(X=2) + a_1 P(X=3) + a_2 P (X=3) + a_3 P(X=3) + \cdots = a_1 \left(\sum_{j=1}^\infty P(X=j) \right) + a_2\left(\sum_{j=2}^\infty P(X=j) \right) + \cdots = a_1 P(X \geq 1) + a_2 P(X \geq 2) + a_3 P(X \geq 3) + \cdots = \sum_{j=1}^\infty a_j P(X \geq j). $$

The key for this problem is the following fact (reference: Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$):

$$ 1 + 2 + \cdots + n = \frac{n(n+1)}{2} = \frac{n^2+n}{2}.$$

To prove this is just a standard induction argument. Notice we can rewrite the above as

$$ 2(1 + 2 + \cdots + n) = 2 + 4 + \cdots + 2n = n^2 + n.$$

Now we use definitions, the first part, and this fact to get

$$E(N(N+1)) = \sum_{j=1}^\infty (j^2 + j) P(X=j) = \sum_{j=1}^\infty \sum_{k=1}^j 2k \cdot P(X=j) = \sum_{j=1}^\infty 2j P(X \geq j) = 2 \sum_{j=1}^\infty j P(X \geq j). $$

User203940
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  • In the last line I'm using this: https://proofwiki.org/wiki/Expectation_of_Function_of_Discrete_Random_Variable – User203940 Oct 16 '20 at 20:02
  • How do you get the first equality in the last line? –  Oct 16 '20 at 20:27
  • That was what my comment was about. $N(N+1) = N^2+N$. Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be defined by $g(x) = x^2 + x$. Then we're evaluating $E(g(N))$, which comes out to $\sum_{k=1}^\infty g(k) P(X=k).$ – User203940 Oct 16 '20 at 20:27