2 Acknowledging mistake

Apologies that this

The Computation Below is not really an answerincorrect, I'm not able to comment yet on but is left up as a learning tool. Please read the sitecomments for details.

Let $X_i$ be i.i.d. uniform r.v.'s on $[-1,1]$. If I compute $P_n$ for $n=3$, I don't get your value. Am I doing something wrong in my computation? I get the following: \begin{align} P_3 & = \mathbb{P}(X_1 > 0, \; X_2+X_1 > 0, \; X_3+X_2+X_1 > 0 ) \newline & = \frac{1}{8}\int_{-1}^1 \int_{-1}^1 \int_{-1}^1\mathbf{1}{\{x>0,\;y>-x,\;z>-(x+y)\}} dz\;dy\;dx \newline &= \frac{1}{8}\int_{0}^1 \int_{-x}^1 \int_{-(x+y)}^1 dz\;dy\;dx \newline &= \frac{1}{8}\int_{0}^1 \int_{-x}^1 (1+x+y) dy\;dx \newline & = \frac{1}{8}\int_{0}^1 (\frac{3}{2}+2x+\frac{x^2}{2}) dx \newline & = \frac{1}{3}. \end{align} This disagrees slightly with your formula, which gives $P_3 = \frac{5}{16}$.

Let me know if I have misunderstood the problem or made an error, and I will remove this immediately. Otherwise, it seems like a possible answer to your question is to generalize this computation by an inductive argument.

1

Apologies that this is not really an answer, I'm not able to comment yet on the site.

Let $X_i$ be i.i.d. uniform r.v.'s on $[-1,1]$. If I compute $P_n$ for $n=3$, I don't get your value. Am I doing something wrong in my computation? I get the following: \begin{align} P_3 & = \mathbb{P}(X_1 > 0, \; X_2+X_1 > 0, \; X_3+X_2+X_1 > 0 ) \newline & = \frac{1}{8}\int_{-1}^1 \int_{-1}^1 \int_{-1}^1\mathbf{1}{\{x>0,\;y>-x,\;z>-(x+y)\}} dz\;dy\;dx \newline &= \frac{1}{8}\int_{0}^1 \int_{-x}^1 \int_{-(x+y)}^1 dz\;dy\;dx \newline &= \frac{1}{8}\int_{0}^1 \int_{-x}^1 (1+x+y) dy\;dx \newline & = \frac{1}{8}\int_{0}^1 (\frac{3}{2}+2x+\frac{x^2}{2}) dx \newline & = \frac{1}{3}. \end{align} This disagrees slightly with your formula, which gives $P_3 = \frac{5}{16}$.

Let me know if I have misunderstood the problem or made an error, and I will remove this immediately. Otherwise, it seems like a possible answer to your question is to generalize this computation by an inductive argument.