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

$$ f_{z}(z) = \sum_{x=0}^{z}\frac{\lambda_{1}^{x}}{x!}e^{-\l…

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

$$ f_{z}(z) = \sum_{x=0}^{z}\frac{\lambda_{1}^{x}}{x!}e^{-\l…