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Showing posts from February, 2013

### The Expected Draws to Sum over One.

Q: You have a random number generator that creates random numbers exponentially between $$[0,1]$$. You draw from this generator and keep adding the result. What is the expected number of draws to get this sum to be greater than 1.

Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)

A: Before getting into the solution for this, I'll go over an established theorem of exponential distributions.

If there exists two random variables which follow a exponential distribution with parameters $$\lambda_1$$ and $$\lambda_2$$ then their sum is given by the convolution of the two probability density functions. This is shown as

$$P(z = X_1 + X_2) = f_{z}(z) = \sum_{x=0}^{z}f_{X_{1}}(x) f_{X_{2}}(z - x)$$

The probability density function of a distribution with rate parameter $$\lambda$$ is given as

$$f(k,\lambda) = \frac{\lambda^{k}e^{-\lambda}}{k!}$$

Plugging this into the convolution formula gives us

### The Case of Two Mariners

Q:Two mariners report to the skipper of a ship that they are distances $$d_1$$ and $$d_2$$ from the shore. The skipper knows from historical data that the mariners A & B make errors that are normally distributed and have a standard deviation of $$s_1$$ and $$s_2$$. What should the skipper do to arrive at the best estimate of how far the ship is from the shore?

A: At a first look, it appears that the simplest solution would be to take the estimate of the navigator who has the lower standard deviation. If $$s_1 < s_2$$ then pick $$d_1$$ else pick $$d_2$$.

But there is a way to do better than that. Assume you take a linearly weighted sum of the two with weight $$= \omega$$.

$$d_{blended} = \omega\times d_1 + ( 1 - \omega)\times d_2$$

The variance of the blended estimate would be given by

$$Var(d_{blended}) = \omega^{2}\times s_{1}^{2} + (1 - \omega)^{2}\times s_{2}^{2}$$

We next proceed to find a value for $$\omega$$ that minimizes the variance $$Var(d_{blended})$$. For this …