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I have found the similar question for Uniform Distribution i.e, U(0,1) but not for Normal distribution,

So basically the problem is, using a standard normal distribution i.e., ∼(,$^2$) random number generator, keep generating and summing until the sum is greater than .

How to find the expected number of moves you will take. I am not even sure the value is defined.

I was modelling the equation like here - Link, and now the equation also has Normal distribution term in the right hand side while picking a number, this converts it to a form where it appears, convolution theorem might be used, tho I am not able to proceed ahead or gain any useful breakthrough.

Welcome to any other methods as well

  • Standard normal distributied random numbers with a mean of $0$ will never sum up to $r$, me think. – m-stgt Feb 28 '24 at 23:45

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Assume $\mu$ and $r$ are positive. Let $Y_i \sim N(\mu, \sigma^2)$ be i.i.d., and let $X_r$ be the minimum number of these variables needed to sum to at least $r$. Then you have $$ E[X_r]=1+\sum_{k=1}^{\infty}P[X_r > k] = 1+\sum_{k=1}^{\infty}P[Y_1+\ldots+Y_k < r]. $$ Since $Y_1+\ldots+Y_k \sim N(k\mu, k\sigma^2)$, this reduces to $$ E[X_r]=1+\sum_{k=1}^{\infty}\Phi\left(\frac{r-k\mu}{\sigma\sqrt{k}}\right), $$ where $\Phi$ is the c.d.f. of the standard normal distribution. This sum converges; the terms (once $k\mu \gg r$) decay exponentially with $k$.

mjqxxxx
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  • How did you write the Probability summation to CDF summation, why the division with sigma root k, also how do we proceed after this, I am not very sure how to proceed with sum where lot of CDF of normal is present – Sunny Chaturvedi Feb 28 '24 at 23:27
  • $P[Y_1+\ldots+Y_k < r]$ is equal to $P[(Y_1 - \mu) + \ldots +(Y_k-\mu) < r - k\mu]$, which in turn is equal to $P[(Y_1 - \mu)/(\sigma\sqrt{k})+ \ldots + (Y_k-\mu)/(\sigma\sqrt{k}) < (r-k\mu)/(\sigma\sqrt{k})]$. But $(Y_i - \mu)/(\sigma\sqrt{k})$ is $\sim N(0, 1/k)$, and so the sum of $k$ such variables is $\sim N(0, 1)$. – mjqxxxx Feb 29 '24 at 00:29