TEST 3 with answers

15 A die is marked with numbers 2,4,8,16,32 and 64. It is rolled twice, the first roll is $X$ and the second $Y$.

1. What is $E(X)$?

2. What is $P(XY < 100)$

3. What is $E(XY)$ (Hint: you can do this without finding the joint density $P(X=j,Y=k)$).

1. $E(X) = 2*(1/6) + 4*(1/6) + 8*(1/6) + 16*(1/6) + 32*(1/6) + 64*(1/6) = 126/6$.

2. Each of the outcomes is equally likely. We need to enumerate the outcomes: If first die is 2 then 2,4,8,16,32 work for second, if first is 4 then 2,4,8,16 work for second ... if first is 64, then none work. Cound them to find 5+4+3+2+1 = 15 such outcomes so the probablity is 15/36

3. Use independence to get $E(XY) = (126/6)^2$.

20 A sequence of independent Bernoulli trials $X_1,X_2,X_3,\dots$ is played with success probability $p$. (For example, roll a dice and count a success if a 6 appears. Then $p=1/6$.)

1. Let $T$ be the time of the first success. What is the distribution of $T$?

2. What is $E(T)$?

3. Let $T_k$ be the time of the $k$ th success. Write $T_k$ as the sum of $k$ independent, identically distributed random variables. Be very specific what these are.

4. Find $E(T_{100})$ and $\text{\textbf{\sf var}}(T_{100})$ using your last answer.

1. $T$ is geometric with parameter p. So $P(T=k) = q^{k-1}p$

2. We know, don’t prove, that $E(T)=1/p$.

3. Let $T_k = S_1+S_2+...+S_k$ where each $S_k$ ios the time between successes. Each has the same distribution as our previous $T$.

4. So $E(T_k) = kE(S_1) = k(1/p)$.

10 Let $X$ be a normal random variable with mean 10 and variance 20.

1. What does Chebyshev’s theorem say about

   $$P(|X - 10| > 40)$$

2. Since $X$ is normal, what is the actual answer. (You shouldn’t need the table for this one.)

1. Chebyshevs says

   $$P(|X - 10| > 40) = P(|X-\mu| > 2 \sigma) \leq 1/2^2 = 1/4$$

2. The normal tells us that only 5% of the observations will be outside 2 standard deviations.

15 Let $X$ be a random variable with probability density

$$f(x) = c x^{-2},\quad 1 < x < 4$$

and 0 otherwise

1. Find the correct value of $c$.

2. Find $E(X)$.

3. Find $\text{\textbf{\sf var}}(X)$.

1. $c$ is found from $\int f(x) dx = 1$. This is

   $$\int_1^4 c x^{-2} dx = c -x^{-1} |^4_1 = c(3/4)$$

   so $c=4/3$.

2. $$E(X) = \int x f(x) dx = \int_1^4 (4/3) x x^{-2} dx = 4/3 \ln x |^4_1 = 4/3(\ln 4 - \ln 1) = 4/3 \ln 4.$$

3. We need to find $E(X^2)$ and then subtract our last answer

   $$E(X^2) = \int x^2 f(x) dx = \int_1^4 (4/3) x^2 x^{-2} dx = (4/3) * (4-1) = 4.$$

   This gives $\text{\textbf{\sf var}}(X) = 4 - (4/3 \ln 4)^2 = 2.9190$.

5 Let $X$ be an exponential random variable with $\lambda=2$. Find the value of $t$ so that $P(X>t)=0.75$.

From the next question, this amounts to solving $P(X > t) = 0.75$ or $e^{-2t} = 3/4$. Take logs to get

$$t = 1/(-2) \ln 3/4 = \ln ( 2 /\sqrt{3}) = 1.2425$$

10 Let $X$ and $Y$ be independent, exponential random variable with parameter $\lambda$. This means $P(X > t) = e^{-\lambda t}$. Let $Z$ be the minumum of $X$ and $Y$.

1. Verify that $Z$ is an exponential random variable. You need to show that $P(Z>t) = e^{-\beta t}$ for some $\beta$.(Hint: use independence somehow.)

2. What is $\beta$.

The key is to observe that if $Z > t$ then both $X>t$ and $Y>t$. Then you can use independece. So

$$P(Z>t) = P(X>t,Y>t) = P(X>t)P(Y>t) = e^{-\lambda t} e^{-\lambda t} = e^{-2 \lambda t}$$

From this you see that $\beta=2\lambda$.