Revisions to How large can $\mathbf{P}[X_1 + X_2 + X_3 < 2 X_4]$ get?

This results in proper spacing to the right of \sim.

Source Link

edited Jul 12 at 22:23

1
12
85
126

I believe the probability is at least $\approx0.343$.

Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$.

Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$. The $Y_i$'s produce the desired inequality with an inclusive $\le$ with probability \begin{align} I+II+III &= P[ \text{binomial}(3, p) \le 1] P[Y_4=1] \\&\quad+ P[ \text{binomial}(3, p) = 2] P[Y_4=1] \\&\quad+ (1-p)^4. \end{align} But the problem requires a strict inequality, for which the second and third terms are lost. We can boost this binomial scheme by using $X_1,X_2,X_3,X_4$ from $\mu_n$, independently $Y_i\sim$$Y_i\sim{}$Bernoulli$(p)$ as above, and set $$Z_i = t X_i + Y_i.$$ for some $t>0$ very small. Assuming that the support of $\mu_n$ is bounded, we can always choose $t>0$ small enough so that the first term is unchanged, while the second term and third terms become strict inequality with independent probability $q_n$: $$ q_{n+1} = P[Z_1+Z_2+Z_3 < 2Z_4] = P[ \text{binomial}(3, p) \le 1] P[Y_4=1] + q_n\Big( P[ \text{binomial}(3, p) = 2] P[Y_4=1] + (1-p)^4\Big). $$ This defines two polynomial $f(p)$ and $g(p)$ of order $4$ in $p\in[0,1]$ such that $$ q_{n+1} = f(p) + q_n g(p). $$ Consider the fixed-point $q_\infty$ defined by $$ q_{\infty} = f(p) + q_{\infty} g(p). $$ We can reach a large $q_\infty$ by maximizing $$ M = \max_{p\in[0,1]} \frac{f(p)}{(1-g(p))_+}. $$ If I am not mistaken, $f(p)= ((1-p)^3 + 3(1-p)^2p )p$ while $g(p) = 3p^2(1-p)p + (1-p)^4$.

The maximum is reached at $p^* \approx 0.404$ giving $M\approx 0.343$.

We now set $p=p^*$ for this maximizer, and perform this procedure, starting from some discrete distribution $q$ (e.g., Bernoulli($p^*$)). Since $g(p^*)\in(0,1)$, the sequence $q_n$ converges to $M\approx 0.343$.

Edit: using Will's explicit formula with $t=0.1$, with $n=30$ terms confirm a probability of around $0.343$:

🖧[14]: import numpy as np
  ...: 
  ...: p = 0.404
  ...: t = 0.1
  ...: powers = np.arange(30)
  ...: ts = (t ** powers)[None, None, :]
  ...: 
  ...: bs = np.random.binomial(1, p, size=(100000, 4, 30))
  ...: ys = np.sum(ts * bs, axis=-1)
  ...: (np.sum(ys[:, :-1], axis=1) <  2 * ys[:, -1]).mean()
🖧[14]: 0.34353

I believe the probability is at least $\approx0.343$.

Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$.

Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$. The $Y_i$'s produce the desired inequality with an inclusive $\le$ with probability \begin{align} I+II+III &= P[ \text{binomial}(3, p) \le 1] P[Y_4=1] \\&\quad+ P[ \text{binomial}(3, p) = 2] P[Y_4=1] \\&\quad+ (1-p)^4. \end{align} But the problem requires a strict inequality, for which the second and third terms are lost. We can boost this binomial scheme by using $X_1,X_2,X_3,X_4$ from $\mu_n$, independently $Y_i\sim$Bernoulli$(p)$ as above, and set $$Z_i = t X_i + Y_i.$$ for some $t>0$ very small. Assuming that the support of $\mu_n$ is bounded, we can always choose $t>0$ small enough so that the first term is unchanged, while the second term and third terms become strict inequality with independent probability $q_n$: $$ q_{n+1} = P[Z_1+Z_2+Z_3 < 2Z_4] = P[ \text{binomial}(3, p) \le 1] P[Y_4=1] + q_n\Big( P[ \text{binomial}(3, p) = 2] P[Y_4=1] + (1-p)^4\Big). $$ This defines two polynomial $f(p)$ and $g(p)$ of order $4$ in $p\in[0,1]$ such that $$ q_{n+1} = f(p) + q_n g(p). $$ Consider the fixed-point $q_\infty$ defined by $$ q_{\infty} = f(p) + q_{\infty} g(p). $$ We can reach a large $q_\infty$ by maximizing $$ M = \max_{p\in[0,1]} \frac{f(p)}{(1-g(p))_+}. $$ If I am not mistaken, $f(p)= ((1-p)^3 + 3(1-p)^2p )p$ while $g(p) = 3p^2(1-p)p + (1-p)^4$.

The maximum is reached at $p^* \approx 0.404$ giving $M\approx 0.343$.

We now set $p=p^*$ for this maximizer, and perform this procedure, starting from some discrete distribution $q$ (e.g., Bernoulli($p^*$)). Since $g(p^*)\in(0,1)$, the sequence $q_n$ converges to $M\approx 0.343$.

Edit: using Will's explicit formula with $t=0.1$, with $n=30$ terms confirm a probability of around $0.343$:

🖧[14]: import numpy as np
  ...: 
  ...: p = 0.404
  ...: t = 0.1
  ...: powers = np.arange(30)
  ...: ts = (t ** powers)[None, None, :]
  ...: 
  ...: bs = np.random.binomial(1, p, size=(100000, 4, 30))
  ...: ys = np.sum(ts * bs, axis=-1)
  ...: (np.sum(ys[:, :-1], axis=1) <  2 * ys[:, -1]).mean()
🖧[14]: 0.34353

I believe the probability is at least $\approx0.343$.

Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$.

Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$. The $Y_i$'s produce the desired inequality with an inclusive $\le$ with probability \begin{align} I+II+III &= P[ \text{binomial}(3, p) \le 1] P[Y_4=1] \\&\quad+ P[ \text{binomial}(3, p) = 2] P[Y_4=1] \\&\quad+ (1-p)^4. \end{align} But the problem requires a strict inequality, for which the second and third terms are lost. We can boost this binomial scheme by using $X_1,X_2,X_3,X_4$ from $\mu_n$, independently $Y_i\sim{}$Bernoulli$(p)$ as above, and set $$Z_i = t X_i + Y_i.$$ for some $t>0$ very small. Assuming that the support of $\mu_n$ is bounded, we can always choose $t>0$ small enough so that the first term is unchanged, while the second term and third terms become strict inequality with independent probability $q_n$: $$ q_{n+1} = P[Z_1+Z_2+Z_3 < 2Z_4] = P[ \text{binomial}(3, p) \le 1] P[Y_4=1] + q_n\Big( P[ \text{binomial}(3, p) = 2] P[Y_4=1] + (1-p)^4\Big). $$ This defines two polynomial $f(p)$ and $g(p)$ of order $4$ in $p\in[0,1]$ such that $$ q_{n+1} = f(p) + q_n g(p). $$ Consider the fixed-point $q_\infty$ defined by $$ q_{\infty} = f(p) + q_{\infty} g(p). $$ We can reach a large $q_\infty$ by maximizing $$ M = \max_{p\in[0,1]} \frac{f(p)}{(1-g(p))_+}. $$ If I am not mistaken, $f(p)= ((1-p)^3 + 3(1-p)^2p )p$ while $g(p) = 3p^2(1-p)p + (1-p)^4$.

The maximum is reached at $p^* \approx 0.404$ giving $M\approx 0.343$.

We now set $p=p^*$ for this maximizer, and perform this procedure, starting from some discrete distribution $q$ (e.g., Bernoulli($p^*$)). Since $g(p^*)\in(0,1)$, the sequence $q_n$ converges to $M\approx 0.343$.

Edit: using Will's explicit formula with $t=0.1$, with $n=30$ terms confirm a probability of around $0.343$:

🖧[14]: import numpy as np
  ...: 
  ...: p = 0.404
  ...: t = 0.1
  ...: powers = np.arange(30)
  ...: ts = (t ** powers)[None, None, :]
  ...: 
  ...: bs = np.random.binomial(1, p, size=(100000, 4, 30))
  ...: ys = np.sum(ts * bs, axis=-1)
  ...: (np.sum(ys[:, :-1], axis=1) <  2 * ys[:, -1]).mean()
🖧[14]: 0.34353

typo/aesthetics

Source Link

edited Jul 12 at 17:30

jlewk

1.7k
7
14

I believe the probability is at least $\approx0.343$.

Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$.

Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$. The $Y_i$'s produce the desired inequality with an inclusive $\le$ haswith probability \begin{align} I+II+III &= P[ \text{binomial}(3, p) \le 1] P[Y_4=1] \\&\quad+ P[ \text{binomial}(3, p) = 2] P[Y_4=1] \\&\quad+ (1-p)^4. \end{align} But the problem requires a strict inequality, for which the second and third terms are lost. We can boost this binomial scheme by using $X_1,X_2,X_3,X_4$ from $\mu_n$, independently $Y_i\sim$Bernoulli$(p)$ as above, and set $$Z_i = t X_i + Y_i.$$ for some $t>0$ very small. Assuming that the support of $\mu_n$ is bounded, we can always choose $t>0$ small enough so that the first term is unchanged, and forwhile the second term and third terms become strict inequality with independent probability $q_n$: $$ q_{n+1} = P[Z_1+Z_2+Z_3 < 2Z_4] = P[ \text{binomial}(3, p) \le 1] P[Y_4=1] + q_n\Big( P[ \text{binomial}(3, p) = 2] P[Y_4=1] + (1-p)^4\Big). $$ This defines two polynomial $f(p)$ and $g(p)$ of order $4$ in $p\in[0,1]$ such that $$ q_{n+1} = f(p) + q_n g(p). $$ Consider the fixed-point $q_\infty$ defined by $$ q_{\infty} = f(p) + q_{\infty} g(p). $$ We can reach a large $q_\infty$ by maximizing $$ M = \max_{p\in[0,1]} \frac{f(p)}{(1-g(p))_+}. $$ If I am not mistaken, $f(p)= ((1-p)^3 + 3(1-p)^2p )p$ while $g(p) = 3p^2(1-p)p + (1-p)^4$.

The maximum is reached at $p^* \approx 0.404$ giving $M\approx 0.343$.

We now set $p=p^*$ for this maximizer, and perform this procedure, starting from some discrete distribution $q$ (e.g., Bernoulli($1/2$$p^*$)). Since $g(p^*)\in(0,1)$, the sequence $q_n$ converges to $M\approx 0.343$.

Edit: using Will's explicit formula with $t=0.1$, with $n=30$ simulations confirmterms confirm a probability of around $0.343$:

🖧[14]: import numpy as np
  ...: 
  ...: p = 0.404
  ...: t = 0.1
  ...: powers = np.arange(30)
  ...: ts = (t ** powers)[None, None, :]
  ...: 
  ...: bs = np.random.binomial(1, p, size=(100000, 4, 30))
  ...: ys = np.sum(ts * bs, axis=-1)
  ...: (np.sum(ys[:, :-1], axis=1) <  2 * ys[:, -1]).mean()
🖧[14]: 0.34353

I believe the probability is at least $\approx0.343$.

Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$.

Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$ The $Y_i$'s produce the desired inequality with an inclusive $\le$ has probability \begin{align} I+II+III &= P[ \text{binomial}(3, p) \le 1] P[Y_4=1] \\&\quad+ P[ \text{binomial}(3, p) = 2] P[Y_4=1] \\&\quad+ (1-p)^4. \end{align} But the problem requires a strict inequality, for which the second and third terms are lost. We can boost this binomial scheme by using $X_1,X_2,X_3,X_4$ from $\mu_n$, independently $Y_i\sim$Bernoulli$(p)$ as above, and set $$Z_i = t X_i + Y_i.$$ for some $t>0$ very small. Assuming that the support of $\mu_n$ is bounded, we can always choose $t>0$ small enough so that the first term is unchanged, and for the second term and third terms become strict inequality with independent probability $q_n$: $$ q_{n+1} = P[Z_1+Z_2+Z_3 < 2Z_4] = P[ \text{binomial}(3, p) \le 1] P[Y_4=1] + q_n\Big( P[ \text{binomial}(3, p) = 2] P[Y_4=1] + (1-p)^4\Big). $$ This defines two polynomial $f(p)$ and $g(p)$ of order $4$ in $p\in[0,1]$ such that $$ q_{n+1} = f(p) + q_n g(p). $$ Consider the fixed-point $q_\infty$ defined by $$ q_{\infty} = f(p) + q_{\infty} g(p). $$ We can reach a large $q_\infty$ by maximizing $$ M = \max_{p\in[0,1]} \frac{f(p)}{(1-g(p))_+}. $$ If I am not mistaken, $f(p)= ((1-p)^3 + 3(1-p)^2p )p$ while $g(p) = 3p^2(1-p)p + (1-p)^4$.

The maximum is reached at $p^* \approx 0.404$ giving $M\approx 0.343$.

We now set $p=p^*$ for this maximizer, and perform this procedure, starting from some discrete distribution $q$ (e.g., Bernoulli($1/2$)). Since $g(p^*)\in(0,1)$, the sequence $q_n$ converges to $M\approx 0.343$.

Edit: using Will's explicit formula with $t=0.1$, with $n=30$ simulations confirm a probability of around $0.343$:

🖧[14]: import numpy as np
  ...: 
  ...: p = 0.404
  ...: t = 0.1
  ...: powers = np.arange(30)
  ...: ts = (t ** powers)[None, None, :]
  ...: 
  ...: bs = np.random.binomial(1, p, size=(100000, 4, 30))
  ...: ys = np.sum(ts * bs, axis=-1)
  ...: (np.sum(ys[:, :-1], axis=1) <  2 * ys[:, -1]).mean()
🖧[14]: 0.34353

I believe the probability is at least $\approx0.343$.

Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$.

Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$. The $Y_i$'s produce the desired inequality with an inclusive $\le$ with probability \begin{align} I+II+III &= P[ \text{binomial}(3, p) \le 1] P[Y_4=1] \\&\quad+ P[ \text{binomial}(3, p) = 2] P[Y_4=1] \\&\quad+ (1-p)^4. \end{align} But the problem requires a strict inequality, for which the second and third terms are lost. We can boost this binomial scheme by using $X_1,X_2,X_3,X_4$ from $\mu_n$, independently $Y_i\sim$Bernoulli$(p)$ as above, and set $$Z_i = t X_i + Y_i.$$ for some $t>0$ very small. Assuming that the support of $\mu_n$ is bounded, we can always choose $t>0$ small enough so that the first term is unchanged, while the second term and third terms become strict inequality with independent probability $q_n$: $$ q_{n+1} = P[Z_1+Z_2+Z_3 < 2Z_4] = P[ \text{binomial}(3, p) \le 1] P[Y_4=1] + q_n\Big( P[ \text{binomial}(3, p) = 2] P[Y_4=1] + (1-p)^4\Big). $$ This defines two polynomial $f(p)$ and $g(p)$ of order $4$ in $p\in[0,1]$ such that $$ q_{n+1} = f(p) + q_n g(p). $$ Consider the fixed-point $q_\infty$ defined by $$ q_{\infty} = f(p) + q_{\infty} g(p). $$ We can reach a large $q_\infty$ by maximizing $$ M = \max_{p\in[0,1]} \frac{f(p)}{(1-g(p))_+}. $$ If I am not mistaken, $f(p)= ((1-p)^3 + 3(1-p)^2p )p$ while $g(p) = 3p^2(1-p)p + (1-p)^4$.

The maximum is reached at $p^* \approx 0.404$ giving $M\approx 0.343$.

We now set $p=p^*$ for this maximizer, and perform this procedure, starting from some discrete distribution $q$ (e.g., Bernoulli($p^*$)). Since $g(p^*)\in(0,1)$, the sequence $q_n$ converges to $M\approx 0.343$.

Edit: using Will's explicit formula with $t=0.1$, with $n=30$ terms confirm a probability of around $0.343$:

🖧[14]: import numpy as np
  ...: 
  ...: p = 0.404
  ...: t = 0.1
  ...: powers = np.arange(30)
  ...: ts = (t ** powers)[None, None, :]
  ...: 
  ...: bs = np.random.binomial(1, p, size=(100000, 4, 30))
  ...: ys = np.sum(ts * bs, axis=-1)
  ...: (np.sum(ys[:, :-1], axis=1) <  2 * ys[:, -1]).mean()
🖧[14]: 0.34353

Add simulations

Source Link

edited Jul 12 at 16:10

jlewk

1.7k
7
14

I believe the probability is at least $\approx0.343$.

Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$.

Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$ The $Y_i$'s produce the desired inequality with an inclusive $\le$ has probability \begin{align} I+II+III &= P[ \text{binomial}(3, p) \le 1] P[Y_4=1] \\&\quad+ P[ \text{binomial}(3, p) = 2] P[Y_4=1] \\&\quad+ (1-p)^4. \end{align} But the problem requires a strict inequality, for which the second and third terms are lost. We can boost this binomial scheme by using $X_1,X_2,X_3,X_4$ from $\mu_n$, independently $Y_i\sim$Bernoulli$(p)$ as above, and set $$Z_i = t X_i + Y_i.$$ for some $t>0$ very small. Assuming that the support of $\mu_n$ is bounded, we can always choose $t>0$ small enough so that the first term is unchanged, and for the second term and third terms become strict inequality with independent probability $q_n$: $$ q_{n+1} = P[Z_1+Z_2+Z_3 < 2Z_4] = P[ \text{binomial}(3, p) \le 1] P[Y_4=1] + q_n\Big( P[ \text{binomial}(3, p) = 2] P[Y_4=1] + (1-p)^4\Big). $$ This defines two polynomial $f(p)$ and $g(p)$ of order $4$ in $p\in[0,1]$ such that $$ q_{n+1} = f(p) + q_n g(p). $$ Consider the fixed-point $q_\infty$ defined by $$ q_{\infty} = f(p) + q_{\infty} g(p). $$ We can reach a large $q_\infty$ by maximizing $$ M = \max_{p\in[0,1]} \frac{f(p)}{(1-g(p))_+}. $$ If I am not mistaken, $f(p)= ((1-p)^3 + 3(1-p)^2p )p$ while $g(p) = 3p^2(1-p)p + (1-p)^4$.

The maximum is reached at $p^* \approx 0.404$ giving $M\approx 0.343$.

We now set $p=p^*$ for this maximizer, and perform this procedure, starting from some discrete distribution $q$ (e.g., Bernoulli($1/2$)). Since $g(p^*)\in(0,1)$, the sequence $q_n$ converges to $M\approx 0.343$.

Edit: using Will's explicit formula with $t=0.1$, with $n=30$ simulations confirm a probability of around $0.343$:

🖧[14]: import numpy as np
  ...: 
  ...: p = 0.404
  ...: t = 0.1
  ...: powers = np.arange(30)
  ...: ts = (t ** powers)[None, None, :]
  ...: 
  ...: bs = np.random.binomial(1, p, size=(100000, 4, 30))
  ...: ys = np.sum(ts * bs, axis=-1)
  ...: (np.sum(ys[:, :-1], axis=1) <  2 * ys[:, -1]).mean()
🖧[14]: 0.34353

I believe the probability is at least $\approx0.343$.

Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$.

Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$ The $Y_i$'s produce the desired inequality with an inclusive $\le$ has probability \begin{align} I+II+III &= P[ \text{binomial}(3, p) \le 1] P[Y_4=1] \\&\quad+ P[ \text{binomial}(3, p) = 2] P[Y_4=1] \\&\quad+ (1-p)^4. \end{align} But the problem requires a strict inequality, for which the second and third terms are lost. We can boost this binomial scheme by using $X_1,X_2,X_3,X_4$ from $\mu_n$, independently $Y_i\sim$Bernoulli$(p)$ as above, and set $$Z_i = t X_i + Y_i.$$ for some $t>0$ very small. Assuming that the support of $\mu_n$ is bounded, we can always choose $t>0$ small enough so that the first term is unchanged, and for the second term and third terms become strict inequality with independent probability $q_n$: $$ q_{n+1} = P[Z_1+Z_2+Z_3 < 2Z_4] = P[ \text{binomial}(3, p) \le 1] P[Y_4=1] + q_n\Big( P[ \text{binomial}(3, p) = 2] P[Y_4=1] + (1-p)^4\Big). $$ This defines two polynomial $f(p)$ and $g(p)$ of order $4$ in $p\in[0,1]$ such that $$ q_{n+1} = f(p) + q_n g(p). $$ Consider the fixed-point $q_\infty$ defined by $$ q_{\infty} = f(p) + q_{\infty} g(p). $$ We can reach a large $q_\infty$ by maximizing $$ M = \max_{p\in[0,1]} \frac{f(p)}{(1-g(p))_+}. $$ If I am not mistaken, $f(p)= ((1-p)^3 + 3(1-p)^2p )p$ while $g(p) = 3p^2(1-p)p + (1-p)^4$.

The maximum is reached at $p^* \approx 0.404$ giving $M\approx 0.343$.

We now set $p=p^*$ for this maximizer, and perform this procedure, starting from some discrete distribution $q$ (e.g., Bernoulli($1/2$)). Since $g(p^*)\in(0,1)$, the sequence $q_n$ converges to $M\approx 0.343$.

I believe the probability is at least $\approx0.343$.

Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$.

Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$ The $Y_i$'s produce the desired inequality with an inclusive $\le$ has probability \begin{align} I+II+III &= P[ \text{binomial}(3, p) \le 1] P[Y_4=1] \\&\quad+ P[ \text{binomial}(3, p) = 2] P[Y_4=1] \\&\quad+ (1-p)^4. \end{align} But the problem requires a strict inequality, for which the second and third terms are lost. We can boost this binomial scheme by using $X_1,X_2,X_3,X_4$ from $\mu_n$, independently $Y_i\sim$Bernoulli$(p)$ as above, and set $$Z_i = t X_i + Y_i.$$ for some $t>0$ very small. Assuming that the support of $\mu_n$ is bounded, we can always choose $t>0$ small enough so that the first term is unchanged, and for the second term and third terms become strict inequality with independent probability $q_n$: $$ q_{n+1} = P[Z_1+Z_2+Z_3 < 2Z_4] = P[ \text{binomial}(3, p) \le 1] P[Y_4=1] + q_n\Big( P[ \text{binomial}(3, p) = 2] P[Y_4=1] + (1-p)^4\Big). $$ This defines two polynomial $f(p)$ and $g(p)$ of order $4$ in $p\in[0,1]$ such that $$ q_{n+1} = f(p) + q_n g(p). $$ Consider the fixed-point $q_\infty$ defined by $$ q_{\infty} = f(p) + q_{\infty} g(p). $$ We can reach a large $q_\infty$ by maximizing $$ M = \max_{p\in[0,1]} \frac{f(p)}{(1-g(p))_+}. $$ If I am not mistaken, $f(p)= ((1-p)^3 + 3(1-p)^2p )p$ while $g(p) = 3p^2(1-p)p + (1-p)^4$.

The maximum is reached at $p^* \approx 0.404$ giving $M\approx 0.343$.

We now set $p=p^*$ for this maximizer, and perform this procedure, starting from some discrete distribution $q$ (e.g., Bernoulli($1/2$)). Since $g(p^*)\in(0,1)$, the sequence $q_n$ converges to $M\approx 0.343$.

Edit: using Will's explicit formula with $t=0.1$, with $n=30$ simulations confirm a probability of around $0.343$:

🖧[14]: import numpy as np
  ...: 
  ...: p = 0.404
  ...: t = 0.1
  ...: powers = np.arange(30)
  ...: ts = (t ** powers)[None, None, :]
  ...: 
  ...: bs = np.random.binomial(1, p, size=(100000, 4, 30))
  ...: ys = np.sum(ts * bs, axis=-1)
  ...: (np.sum(ys[:, :-1], axis=1) <  2 * ys[:, -1]).mean()
🖧[14]: 0.34353

typo where t is located in the weighted sum

Source Link

edited Jul 12 at 15:55

jlewk

1.7k
7
14

Loading

Source Link

created Jul 12 at 15:44

jlewk

1.7k
7
14

Loading

Stack Exchange Network

Return to Answer