## I discover vegetables taste good in Dijon sauce then proceed to spread the gospel

I had a dozen eggs in the fridge that I need to use up by Wednesday; now I’m down to four. I made Jaffa cakes last week (not bad, but not memorable, and definitely not worth the effort: my main complaint is that the cake portion is bland and has a too-tough texture) and I left some chocolate blueberry cookie dough sitting in the fridge overnight which I’ll bake up tonight. That accounted for four of the eggs. This post is about the two meals that account for another four, and maybe the remaining four also 🙂

I’m in love with this dish because it’s super simple to prepare, filling, and not too unhealthy: steak, eggs, and steamed veggies in a Dijon sauce. Surprisingly, considering how much I love meat, the star of the show here is the veggies in sauce: steamed baby carrots and French green beans (haricot verts) with an emulsion of Dijon mustard, red wine vinegar, EVOO, and some salt. Basically, it’s a quick and dirty substitute for Sauce Dijon.

If I’d known sauce + steamed vegetables was such a winning combination (don’t ask me why it took so late in life for me to get a clue) I would be a much healthier person 🙂 My mission for next week is to learn how to make Hollandaise sauce and its derivatives, and find other vegetables to smother in sauce. No doubt Alton Brown has something interesting to say on the matter.

## Puzzled by the definition of stationarity

I’m reading Santosh Vempala’s survey “Geometric Random Walks: A Survey,” and already I’m puzzled at one of the very first definitions he gives.

Define a Markov chain as a state space sigma algebra pair $$(K, \mathcal{A})$$ with transition probability measures given by $$P_u$$ for each $$u \in K.$$

A distribution $$Q$$ on $$(K, \mathcal{A})$$ is called stationary if one step from it gives the same distribution, i.e., for any $$A \in \mathcal{A},$$
$\int_A P_u(A) \, dQ(u) = Q(A).$

This definition makes sense in words, but mathematically it doesn’t seem sound: unless $$P_u(A) = 1$$ a.e. (with respect to $$u \in A$$), this equality can’t hold. I must be missing something…

Update:
The definition should involve integration over the whole space:

A distribution $$Q$$ on $$(K, \mathcal{A})$$ is called stationary if one step from it gives the same distribution, i.e., for any $$A \in \mathcal{A},$$
$\int_K P_u(A) \, dQ(u) = Q(A).$

That this is so can be seen from the discrete case. Just goes to show you that you have to be careful even when reading peer-reviewed articles.