# Show median zero and symmetry of differences of Gumbel [closed]

Consider a random vector $\epsilon\equiv (\epsilon_1, \epsilon_2, \epsilon_0)$. Suppose $\epsilon_1, \epsilon_2, \epsilon_0$ are i.i.d., distributed as Gumbel with location $0$ and scale $1$.

Take $V\equiv (V_1, V_2, V_3)$ with $$V_1\equiv \epsilon_1-\epsilon_0$$ $$V_2\equiv \epsilon_2-\epsilon_0$$ $$V_3\equiv V_1-V_2$$

Could you help me to show - if true - that $\forall j \in \{1,2,3\}$

1) The distribution of $V_j$ has median $0$

2) The distribution of $V_j$ is symmetric around zero

## closed as off-topic by heropup, Mostafa Ayaz, Namaste, Isaac Browne, LeucippusAug 1 '18 at 2:16

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## 2 Answers

The Gumbel distribution is not important here. In general, whenever $X$ and $Y$ have the same distribution and are independent, then $X-Y$ will be symmetric about zero. This is because $X-Y$ has the same joint distribution as $Y-X$, so for all $c\in \mathbb R$, $$P(X-Y\le c)=P(Y-X\le -c)\stackrel{X-Y\stackrel{d}=Y-X}=P(X-Y\le -c)$$ which says $X-Y$ is symmetric about $0$. This immediately implies its median of $X-Y$ is $0$, since $P(X-Y\le 0)=P(X-Y\ge 0$) and these probabilities sum to $1+P(X-Y=0)\ge 1$, so they must both be at least $\frac12$.

Use this property: if $X$ and $Y$ are distributed as Gumbel distribution $\text{Gum}(\alpha, \beta)$ then $X - Y \sim \text{Log}(0, \beta)$ (see Logistic distribution) and since for logistic distribution has median equal to location parameter $\mu = 0$ then the $\text{med} V_j = 0$. Logistic distribution is symmetric as well, so the second part of your question is also true.