2つの確率変数の合計としての一様確率変数

GrimmetおよびStirzakerから取得：

そうでないことを示すU = X + Y Uは均一[0,1]上に分散され、XおよびYは独立しており、同一分布。あなたはないはず XとYが連続変数であることを前提としています。$U=X+Y$$U$$X$$Y$

場合の矛盾で十分によって簡単証明$X$、$Y$それが常に可能見つけることと主張することによって別個に仮定され$u$及び$u'$その結果、$P(U\leq u+u') \geq P(U\leq u)$一方、$P(X+Y \leq u) = P(X+Y \leq u+u')$。

ただし、この証明はX には拡張されず、$X,Y$は完全に連続または特異連続です。ヒント/コメント/批評？

結果は写真で証明できます。目に見える灰色の領域は、均一な分布は2つの独立した同一に分布した変数の合計として分解できないことを示しています。

表記法

ましょX及びYは、そのようなことはIIDであるX + Yが上に均一な分布を有する[ 0 、1 ]。すべてのそれはこの手段0 ≤ ≤ B ≤ 1、$X$$Y$$X+Y$$[0,1]$$0\le a \le b \le 1$

共通の分布の本質的な支持X及びY故には[ 0 、1 / 2 ]（別段のためにその正の確率が存在することになるX + Yの外側にある[ 0 、1 ]）。$X$$Y$$[0,1/2]$$X+Y$$[0,1]$

絵

してみましょう0 < ε < 1 / 4。ランダム変数の合計がどのように計算されるかを示すこの図を考えてください：$0 \lt \epsilon \lt 1/4$

基礎となる確率分布は、（X 、Y ）の結合分布です。任意のイベントの確率は< X + Y ≤ Bは線間延伸対角線バンドによってカバーされる全確率で与えられるX + Y =およびX + Y = B。このような3つのバンドが示されている：から0までε、左下に小さな青い三角形として現れます。1 / 2 - εに1 /$(X,Y)$$a \lt X+Y \le b$$x+y=a$$x+y=b$$0$$\epsilon$$1/2-\epsilon$ +ϵ、2つの（黄色と緑の）三角形で覆われた灰色の長方形として表示されます。および 1 − ϵから 1で、右上に小さな赤い三角形として表示されます。$1/2+\epsilon$$1-\epsilon$$1$

写真が示すもの

図の左下の三角形をそれを含む左下の正方形と比較し、XおよびYの iid仮定を活用することにより、$X$$Y$

Note that the inequality is strict: equality is not possible because there is some positive probability that both $X$ and $Y$ are less than $\epsilon$ but nevertheless $X+Y \gt \epsilon$.

Similarly, comparing the red triangle to the square in the upper right corner,

最後に、左上と右下にある2つの反対側の三角形を、それらを含む対角バンドと比較すると、別の厳密な不等式が得られます。

The first inequality ensues from the previous two (take their square roots and multiply them) while the second one describes the (strict) inclusion of the triangles within the band and the last equality expresses the uniformity of $X+Y$. The conclusion that $2\epsilon \lt 2\epsilon$ is the contradiction proving such $X$ and $Y$ cannot exist, QED.

I tried finding a proof without considering characteristic functions. Excess kurtosis does the trick. Here's the two-line answer: $\text{Kurt}(U) = \text{Kurt}(X + Y) = \text{Kurt}(X) / 2$ since $X$ and $Y$ are iid. Then $\text{Kurt}(U) = -1.2$ implies $\text{Kurt}(X) = -2.4$ which is a contradiction as $\text{Kurt}(X) \geq -2$ for any random variable.

Rather more interesting is the line of reasoning that got me to that point. $X$ (and $Y$) must be bounded between 0 and 0.5 - that much is obvious, but helpfully means that its moments and central moments exist. Let's start by considering the mean and variance: $\mathbb{E}(U)=0.5$ and $\text{Var}(U)=\frac{1}{12}$. If $X$ and $Y$ are identically distributed then we have:

So $\mathbb{E}(X) = 0.25$. For the variance we additionally need to use independence to apply:

Hence $\text{Var}(X) = \frac{1}{24}$ and $\sigma_X = \frac{1}{2\sqrt{6}} \approx 0.204$. Wow! That is a lot of variation for a random variable whose support ranges from 0 to 0.5. But we should have expected that, since the standard deviation isn't going to scale in the same way that the mean did.

Now, what's the largest standard deviation that a random variable can have if the smallest value it can take is 0, the largest value it can take is 0.5, and the mean is 0.25? Collecting all the probability at two point masses on the extremes, 0.25 away from the mean, would clearly give a standard deviation of 0.25. So our $\sigma_X$ is large but not impossible. (I hoped to show that this implied too much probability lay in the tails for $X + Y$ to be uniform, but I couldn't get anywhere with that on the back of an envelope.)

Second moment considerations almost put an impossible constraint on $X$ so let's consider higher moments. What about Pearson's moment coefficient of skewness, $\gamma_1 = \frac{\mathbb{E}(X - \mu_X)^3}{\sigma_X^3} = \frac{\kappa_3}{\kappa_2^{3/2}}$? This exists since the central moments exist and $\sigma_X \neq 0$. It is helpful to know some properties of the cumulants, in particular applying independence and then identical distribution gives:

This additivity property is precisely the generalisation of how we dealt with the mean and variance above - indeed, the first and second cumulants are just $\kappa_1 = \mu$ and $\kappa_2 = \sigma^2$.

Then $\kappa_3(U) = 2\kappa_3(X)$ and $\big(\kappa_2(U)\big)^{3/2} = \big(2\kappa_2(X)\big)^{3/2} = 2^{3/2} \big(\kappa_2(X)\big)^{3/2}$. The fraction for $\gamma_1$ cancels to yield $\text{Skew}(U) = \text{Skew}(X + Y) = \text{Skew}(X) / \sqrt{2}$. Since the uniform distribution has zero skewness, so does $X$, but I can't see how a contradiction arises from this restriction.

So instead, let's try the excess kurtosis, $\gamma_2 = \frac{\kappa_4}{\kappa_2^2} = \frac{\mathbb{E}(X - \mu_X)^4}{\sigma_X^4} - 3$. By a similar argument (this question is self-study, so try it!), we can show this exists and obeys:

The uniform distribution has excess kurtosis $-1.2$ so we require $X$ to have excess kurtosis $-2.4$. But the smallest possible excess kurtosis is $-2$, which is achieved by the $\text{Binomial}(1, \frac{1}{2})$ Bernoulli distribution.