1 Exponential random variables

Suppose I have two types of light bulbs, type-ex and type-wye, whose lifetimes in years are given by exponential random variables \(X\) and \(Y\), such that \(\mathbb{P}(X >x) = e^{-x/2}\) for all \(x \geq 0\), and \(\mathbb{P}(Y >y) = e^{-y/3}\) for all \(y \geq 0\).

Suppose I install one type-ex light bulb and one type-wye light bulb, and the light bulbs function independently of one another. What is the probability that the ex light bulb dies before the wye light bulb?

2 Grouping coins

Suppose I flip a fair coin 20 times. Assume that the coin flips are independent. Show that the probability that I get a consecutive run of at least four heads is greater than \(0.27\). Hint: group the coins into 5 blocks.

3 Coupon collectors

Suppose that when Tessa buys a Kinder surprise candy egg to get the Kinder surprise toy. Each egg contains one of \(n\) toys uniformly at random, independently of each other. She wishes to collect all \(n\) toys. Let \(T\) be the number of toys her father has to buy in order for her to collect all \(n\) toys. Compute \(\mathbb{E} T\). You should obtain a nice formula that will allow you to compute easily \(\mathbb{E} T\) if \(n=7\). Hint: let \(T_i\) be the number of toys that need to be purchased, after collecting \(i-1\) distinct toys, then \[ T = T_1 + \cdots +T_n.\]

4 Solutions

4.1 Exponential random variables

Since \(X\) and \(Y\) are independent and their pdfs are given by \(f_X(x) = 1/2e^{-x/2}\) and \(f_Y(y) = 1/3e^{-y/3}\), when \(x,y \geq 0\), then the joint pdf is simply given by the product \(f_Xf_Y\); all the probabilities to do with \((X,Y)\) can be found by integration. Let \(R = \{(x,y): 0\leq x <y\}\). We compute

\[\begin{eqnarray*} \mathbb{P}(X <Y) &=& \int\int_R (\frac{1}{2})e^{-x/2}(\frac{1}{3})e^{-y/3}dxdy \\ &=& \int_0 ^{\infty} \frac{1}{2}e^{-x/2}\Big( \int_x^{\infty}\frac{1}{3}e^{-y/3}dy \Big) dx \\ &=& \int_0 ^{\infty} \frac{1}{2}e^{-x/2} \Big[ -e^{-y/3} \Big]_x^{\infty} dx \\ &=& \int_0 ^{\infty} \frac{1}{2}e^{-x/2}e^{-x/3} dx \\ %&=& \int_0 ^{\infty} \frac{1}{2}e^{-x/2}e^{-x/3} dx \\ &=& \int_0 ^{\infty} \frac{1}{2}e^{-5x/6} dx \\ &=& \Big[ -\frac{3}{5} e^{-5x/6} \Big]_0 ^{\infty} \\ &=& 3/5. \end{eqnarray*}\]

4.2 Grouping coins

Let \(F\) be the event that there is a run of at least four heads. Partition the coin-flips into 5 blocks of 4. Consider the event \(E\) of getting four consecutive heads within at least one of the blocks. Clearly, \(E \subseteq F\). The probability of not getting four heads in a block is \(1-(1/2)^4\). So \(\mathbb{P}(E^c) =(1-(1/2)^4)^5\) and \(\mathbb{P}(E) = 1- \mathbb{P}(E^c) \approx 0.275803.\)

4.3 Coupon collector

Note that \(T_1=1\), since at the start, any toy is new. For the next toy, any of the \(n-1\) remaining ones will do; if we get the one we first got, we have to try again. Thus the probability of success is \((n-1)/n\) and failure \(1/n\), and \(T_2\) is geometric with mean \(n/(n-1)\). Similarly, \(T_i\) is geometric with mean \(n/(n-i +1)\), and hence the linearity of expectation gives \[\begin{eqnarray*} \mathbb{E} T &=& \sum_{i=1} ^n \frac{n}{n-i+1} \\ &=& n \sum_{i=1}^n \frac{1}{n-i+1} \\ &=& n \sum_{i=1}^n \frac{1}{i}. \end{eqnarray*}\]

See also

4.4 Convolution

We recall that the pdf for the sum of two independent continuous random variables with pdfs \(f\) and \(g\) is given by a convolution formula:

\[z \mapsto (f \star g)(z) = \int f(z-x)g(x)dx.\]

In our case of two uniforms, the sum lies in \([0,2]\). Thus if \(0\leq z <1\), we have

\[ z \mapsto \int_0 ^ z 1 \cdot 1 dx = z;\]

and if \(1 \leq z \leq 2\), we have \[ z \mapsto \int_{z-1} ^ 1 1 \cdot 1 dx = 2-z.\]

Plotting this function of \(z\) gives a triangle!

5 Version: 05 October 2021