Cannabis use disorder and subsequent risk of depression and bipolar disorder

Post by m0llusk » Sun, Sep 17, 2023, 12:25 AM UTC

Eddy_Viscosity2 wrote: post #283

m0llusk wrote: post #205

This is potentially interesting, but it is important to bear in mind just how tricky mental disorders are to characterize robustly. Psychiatry does not have a record of strongly reproducible results. All of these primary criteria, cannabis use disorder, bipolar disorder, and depression are frequently diagnosed differently by practiced professionals. One of the more academic investigations of addiction, the book High…

> the very concept of "cannabis use disorder" may not really make sense. I've known people who used cannabis the way alcoholics use alcohol - if it wasn't a use disorder, I don't know what is. The concept is valid and can apply to sorts of things from gambling, pron, online gaming even. If the use becomes problematic then its a problem, 'use disorder' is the just the name for this.

Post by Kerb_ » Sun, Sep 17, 2023, 7:43 PM UTC

no_butterscotch wrote: post #219

ajfdsaaj4e4ea wrote: post #83

I'm terrified of seeking help for mental illness because certain diagnosis can have legal impacts. It is a sad truth, no one has a real response for it except "get over it", and it is very, very difficult to get support for political reforms to protect the mentally ill. I am also terrified the medications we force on some mental health patients, especially injected long-term ones, are purely for the benefit of the ca…

> I'm terrified of seeking help for mental illness because certain diagnosis can have legal impacts. It is a sad truth, no one has a real response for it except "get over it", and it is very, very difficult to get support for political reforms to protect the mentally ill. What are some of the impacts? I remember the show "Homeland" the main character had a sister who was a doc who would prescribe her meds for bipolar…

Post by bumby » Tue, Sep 19, 2023, 3:05 AM UTC

tchaffee wrote: post #311

bumby wrote: post #304

Earlier quoted context omitted.

That's just it, though. The WHO says there is no level of alcohol use that is "zero risk." The difference between toxicity and therapeutic use is what helps define the safety of the substance. There are some, like psilocybin mushrooms, that have a relatively monstrous gulf between therapeutic doses (in the mcg) and toxic doses (presumably in the kg). They are quite safe, though not zero risk. I believe alcohol is rel…

They drink wine on a daily basis in at least some of the blue zones where people regularly live to a hundred years old or longer. So it's not reasonable to say alcohol is relatively unsafe. Relative to what and in which circumstances? As far as cannabis being "pretty darn safe", a recent study shows it causes heart disease. And more than one study shows it increases the risk of heart attack for up to an hour. What we…

Post by avodonosov » Sun, Sep 24, 2023, 2:02 PM UTC

avodonosov wrote: post #327

avodonosov wrote: post #84

Earlier quoted context omitted.

I hoped someone who knows statistics will just tell us, instead of me stretching my brain. The famous phrase "correlation does not mean casuation" is not equal to "correlation is indistingushable from casuation", but that's how people seem to often treat it. Ideally, the paper abstract should be clear about what can or can not be concluded. In the title they say "subsequent" as if suggesting casuality, but in the abs…

He-he, the sides of inequality in my formula are always equal :) (c&d/c)/(d/total) > (c&d/d)/(c/total) c&d*total/c*d > c&d*total/d*c

    function takeMeasurement() {
      // c (the cause) itself has probability 0.5
      // d (the dependent) in absence of c has probability 0.5
      //   but the presence of c increases it to 0.7
      const c = Math.random(1.0) 

    P(c) = 0.5
    P(d|!c) = 0.5
    P(d|c) = 0.7
    
    P(d) = ?
    P(c|!d) = ?
    P(c|d) = ?

    -------------------------------

    P(d) = P(c)*P(d|c) + P(!c)*P(d|!c) =
      = P(c)*P(d|c) + (1 - P(c))*P(d|!c) =
      = 0.5*0.7 + 0.5*0.5 = 0.35 + 0.25 =
      = 0.6

    P(c|!d) = P(c) * P(!d|c) / P(!d) =
      = P(c) * (1 - P(d|c)) / (1 - P(d)) =
      = 0.5 * 0.3 / 0.4 =
      = 0.374(9)
    
    P(c|d) = P(c) * P(d|c) / P(d) =
      = 0.5 * 0.7 / 0.6 = 0.35 / 0.6 =
      = 0.58(3)
    
    -------------------------------
    Bayes formula:
    
    P(a|b) = P(a) * P(b|a) / P(b)

    function takeMeasurement2() {
      const d = Math.random(1.0)