I downloaded the surface PM 2.5 data and called up the AI assistant to create a shiny App. I took quite a few prompts and some crucial suggestions (do the h3 grids etc) to get the final app. The end result is pretty decent. Hosted it here using Posit Connect. The github repo is here

The Posit AI is giving USD 5 of free usage in trial and I burned through it pretty fast on making this app and I would like to polish it further. Not sure, how much good the $20 per month would be for multiple projects.

I have got these numbers from Groww for easy representation

The statement above made by the influencer is clearly wrong and see if you could identify the fallacy.

If the statement is true, then there is an arbitrage opportunity to make free money, take a 10lakh loan at 9% and invest in bank FD at 8% and make money at the end ?

The interest rate in the (car) loan is for the time value of money plus the risk premium, sure the interest component of EMI reduces as you pay down your loan every month. But, is it correct to say you would be better off taking the loan at a higher rate (than what you would get in a FD) and investing the 10lakh in FD ? for a moment forget about tax on FD. The FD’s future value of INR 14.85 lakh after 5 years means that it is equal to today’s 10lakh at 8% annual discount rate ¹. You can see that the FD amount is paid to the investor in full only after the term, not during the term. So, if we want to compare the loan, it is only fair that you compare the payments in the same terms.

Sharing slides of my useR!2025 talk, CSV to Parquet: Managing Data for Multi-language Analytics Teams

We can do a test of proportions, and our test statistic is (-)/

The sample size (n) is 1000 and we have observed heads 400 times meaning = 0.4 and = 0.6.

The sample mean is 400 (1000 x 0.4) (X is binomial)

here P = Q = 0.5 (population estimate)

Now the denominator is the standard error of the sample proportion

This variance scaling is not really intuitive at the first look.

The variance is scaled by square of n, so the variance of sample proportion is which is PQ/n and hence the standard error (same as standard deviation) is

We could have solved the problem by testing number of heads instead of proportion of heads (400-500/15.81). 15.81 is just simply

This seemingly simple problem statement has tortured novices and pros alike. To answer this question, we need to know how the odds change for winning if we switch vs if we don’t. When we don’t switch the probability of winning would remain at 1/3. When we switch there is 1/3 chance of selecting a car and losing and 2/3 chance of choosing a goat and winning (due to switch), so switching nearly doubles our chances.Thank you for the extra 33.3%. ¹

I have simulated this scenario below in R

library(dplyr)

set.seed(1)

  doors = c('A', 'B', 'C')
  car_door = sample(doors, size = 10000, replace = T)
  choosen_door = sample(doors, size = 10000, replace = T)

set.seed(1)

tibble(
       car = car_door,
       choosen = choosen_door) |>
  rowwise() |>
  mutate(door_revealed = sample(setdiff(doors, c(car, choosen)), 1)
         ) |>
  mutate(switch = setdiff(doors, c(choosen, door_revealed))) |>
  ungroup()|>
  mutate(remain = choosen) %>% 
  mutate(switch_win = ifelse(car == switch, 1, 0),
         remain_win = ifelse(car == remain, 1, 0)
         ) |>
  summarize(
    switch_win_rate = sum(switch_win)/n(),
    remain_win_rate = sum(remain_win)/n()
  )

# A tibble: 1 × 2
  switch_win_rate remain_win_rate
            <dbl>           <dbl>
1           0.665           0.335

1. Albert Rapp
2. Quarto on Github Pages
3. Building with Quarto - Robert Mitchell
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