Comment by khafra

14 years ago

A physics degree can get you a mid-six-figure job as a quant on Wall Street.

Does "mid-six-figure" mean $550k (midpoint of the six-figure range), $300k (ditto but using the geometric mean), or $150k (midpoint of the $100k-$200k range)?

(My impression is that $150k is a plausible entry-level quant salary, but it seems a bit strange to describe that as "mid-six-figure".)

  • The "x-figure" notation is essentially logarithmic with an offset: "6-figures" = 10^5. Therefore, "mid-six-figures" should be roughly 10^5.5 or $316227.

    • I believe that's the geometric mean he's referring to.

        (a*b)^1/2 # geometric mean, square root of the product of the two numbers
        (10^5 * 10^6)^(1/2) # substitute in for 10^5 and 10^6, the limits of the six-figure range
        (10^(5+6))^(1/2) # simplify inner product
        (10^11)^(1/2) # more simplify
        10^(11 * 1/2) # move the 1/2 in, and ...
        10^(11/2) # simplify
        10^5.5 # and finally you get the same result
      

      The geometric mean always struct me as an interesting idea in mathematics

  • Curious: how could $150k be mid six-figure? I can accept the roughly 500K or 300K meanings, but if someone told me mid six figure, and they meant roughly $150K, I'd feel like they deliberately mislead or lied about the salary. There's no way I can see a reasonable person expecting $150K when someone is referring to "mid six-figure".