<p>Pick any set of non-negative <span class="math-container">$p_1,\dots,p_n$</span> such that <span class="math-container">$p_1+\cdots+p_n=1$</span>. The convex combination of degenerate lotteries <span class="math-container">$L^1,\dots,L^n$</span> with the <span class="math-container">$p_i$</span>'s can be written as 
<span class="math-container">\begin{align}
p_1L^1+\cdots+p_nL^n&amp;=p_1(1,0,\dots,0)+\cdots+p_n(0,\dots,0,1)\\
&amp;=\bigl(p_1(1)+p_2(0)+\cdots+p_n(0),\;\dots,\;p_1(0)+\cdots+p_{n-1}(0)+p_n(1)\bigr)\\
&amp;=(p_1,\dots,p_n).
\end{align}</span>
This is just a lottery <span class="math-container">$L=(p_1,\dots,p_n)$</span>.</p>

CC BY-SA 4.0
2019-11-05T17:26:56.810Z
2019-11-05T17:26:56.810Z
Herr K.
<p>I know that a degenerate lottery is a lottery that yields outcome <span class="math-container">$n$</span> with probability <span class="math-container">$1$</span> and I also know the definition of convex combination: given <span class="math-container">$x_{1},x_{2}, \cdots ,x_{n} \in \mathbb{R}$</span>, a convex combination of these points is a point of the form <span class="math-container">$\alpha_{1}x_{1}+...+\alpha_{n}x_{n}$</span> such that <span class="math-container">$\alpha_1+...+\alpha_n=1$</span>.</p>

<p>But I am quiet confused about why we can write any <span class="math-container">$L=(p_1,...,p_n)$</span> as a convex combination of the degenerate lotteries <span class="math-container">$(L^1,...,L^n)$</span>. Can someone please explain this for me? Many thanks!</p>

<expected-utility>
Why can we write any lottery as a convex combination of the degenerate lotteries?
Christopher Hernandez
# Why can we write any lottery as a convex combination of the degenerate lotteries?

(Asked by: username_0 on 2019-11-05)

I know that a degenerate lottery is a lottery that yields outcome $n$ with probability $1$ and I also know the definition of convex combination: given $x\_{1},x\_{2}, \cdots ,x\_{n} \in \mathbb{R}$, a convex combination of these points is a point of the form $\alpha\_{1}x\_{1}+...+\alpha\_{n}x\_{n}$ such that $\alpha\_1+...+\alpha\_n=1$.

But I am quiet confused about why we can write any $L=(p\_1,...,p\_n)$ as a convex combination of the degenerate lotteries $(L^1,...,L^n)$. Can someone please explain this for me? Many thanks!

**Tags:** expected-utility

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## 1 answers

### Answer by username_1 on 2019-11-05. Score: 4

Pick any set of non-negative $p\_1,\dots,p\_n$ such that $p\_1+\cdots+p\_n=1$. The convex combination of degenerate lotteries $L^1,\dots,L^n$ with the $p\_i$'s can be written as
\begin{align}
p\_1L^1+\cdots+p\_nL^n&=p\_1(1,0,\dots,0)+\cdots+p\_n(0,\dots,0,1)\\
&=\bigl(p\_1(1)+p\_2(0)+\cdots+p\_n(0),\;\dots,\;p\_1(0)+\cdots+p\_{n-1}(0)+p\_n(1)\bigr)\\
&=(p\_1,\dots,p\_n).
\end{align}
This is just a lottery $L=(p\_1,\dots,p\_n)$.


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