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Results tagged with self-study
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user 40036
A routine exercise designed to test one's knowledge; often from a textbook, course, or test used for a class or self-study. This community's policy is to "provide helpful hints" for such questions rather than complete answers.
14
votes
Accepted
Probability of each of the three Christmas puddings having exactly 2 coins
Call the number of pieces in each section $A$, $B$, and $C$. Because $A+B+C=6$, you are interested in $Pr(A=2, B=2) = Pr(B=2|A=2)Pr(A=2)$.
$Pr(A=2)$ is a simple binomial calculation: $A\sim Binom(6, …
- 4,389
12
votes
Accepted
How is it possible to get a high $R^²$ & still have 'poor predictions'?
As you state, $R^2 = 1-SSE/SST$, where SSE is the sum of squared residuals of the model and SST is the sum of squared residuals of a simple model that just predicts the average response variable for e …
- 4,389
2
votes
Accepted
Non-monotone Survival Function?
You have hazard rate
$$
\lambda(t) = 0.2(1+\sin(t\pi/12))
$$
The survival function $S(y)$ -- the proportion surviving through time $y$ -- is
$$
S(y) = \exp\bigg(-\int_0^y \lambda(t) dt\bigg) \\
$$
…
- 4,389
2
votes
Accepted
Expressing a sum as a compound Poisson Distribution
I assume $X_1$ and $X_2$ are independent.
First, here's some intuition: whenever you get an arrival in process $X_1$ (rate 1), your value of $Y$ goes down by 3. Whenever you get an arrival in process …
- 4,389
1
vote
How to do the Proof of P(A ∪ (B ∩ C))?
One way to approach this algebraically is to note that $Pr(a) = E[A]$, where $A$ is a random variable that takes value 1 when event a occurs and 0 when event a does not occur. Logical expressions with …
- 4,389
1
vote
Accepted
Expected time to get all four unique coupons
One way to approach this would be to break down $X$, the number of trials required, into the sum $X = X_1 + X_2 + X_3 + X_4$, where $X_i$ is the number of trials needed to get the $i^{th}$ unique coup …
- 4,389
1
vote
Accepted
$n^{-1}\sum_{i=1}^{n}(1-R_i)\overset{p}{\to}P(R_i=0)$, where $R_i$ is a binary variable
Assume $R_1, R_2, \ldots, R_n$ are IID Bernoulli random variables. By the Weak Law of Large Numbers, $n^{-1}\sum_{i=1}^n (1-R_i)$ converges in probability to $\mathbb{E}[1-R_i] = 1 - \mathbb{E}[R_i] = …
- 4,389