1

The question is:

Show if X is a constant function, then it is a random variable with respect to any sigma-algebra.

With reference to the following links that have already answered this question:

I am seeking confirmation that my understanding is correct with a clear-cut worked example (which would further enhance the understanding of the above links for anyone in the future).

The quotes below are taken from Link 3

Lets assume the stock price is 4 and a coin will be tossed twice, based on the outcome of the tosses, the stock price (S) will be either:

  • $S(\omega _{HH}) = 16$;
  • $S(\omega _{HT}) = 4$;
  • $S(\omega _{TH}) = 4$;
  • $S(\omega _{TT}) = 1$;

Therefore $\Omega =\left \{ \omega _{HH}, \omega _{HT},\omega _{TH},\omega _{TT} \right \}$.

If f==c is constant it is ALWAYS measurable (for any sigma-algebra). This holds as f^{-1}[C] is X if c is in C and empty if c is not in C. And both sets are in any sigma-algebra.

At time zero, we have no idea what the coin tosses will be. For the random variable $S_{0}(\omega)$, our best guess is the current stock price of 4 for all $\omega$ $\epsilon$ $\Omega$. The sigma-algebra is just the trivial set been $F_{0} = \left \{ \varnothing ,\Omega \right \}$.

As the next sigma-algebra or filtration $F_{1}$ will contain more information including $F_{0}$, then the constant function $S_{0}(\omega)=4$ is a random variable with respect to any sigma algebra as as any other sigma-algebra will also contain $\left \{ \varnothing ,\Omega \right \}$.

On the other hand, if f is F-measurable and non-constant, then it assumes at least two values c1 and c2. The set f^{-1}[{c1}] must be in F (by being F-measurable, as {c1} is a closed set) but this set is non-empty (as c1 IS a value of f)...

After the first coin toss the information known is $F_{1} = \left \{ \varnothing ,\Omega,\left \{ HH,HT \right \},\left \{TT,TH \right \} \right \}$ and $S_{1}(\omega)$ is not constant as:

  • $S_{1}\left ( HH,HT \right ) =8$
  • $S_{1}\left ( TT,TH \right ) =2$

...and not X (as the points x where f assumes the value c2 are not in it). So this set cannot be in F, and so f must be constant.

I am a little confused by this. I think it means as the sets $\left \{HH,HT \right \}, \left \{TT,TH \right \}$ do not result in a constant $S_{1}(\omega)$ they can not be in $F_{0}$, therefore only the random variable $S_{0}(\omega)$ is constant?

Tejay Lovelock
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  • @Math1000 just drawing your attention to this question as you were very helpful in answering the questions in link 1 and 2 above. Thanks. – Tejay Lovelock May 23 '17 at 12:38
  • Sorry, but this looks quite confused. You define a random variable $S$, but then go on to talk about two other random variables $S_0$ and $S_1$. How are those random variables defined? What do you even mean by $S_1(HH, HT)=8$, for example? – aduh May 23 '17 at 16:59
  • @aduh thanks for the reply. It's the same random variable S, S_1 indicates there has been 1 coin toss. After the 1st coin toss, S can be either 8 or 2. Likewise at time 0, there have been no coin tosses, so S_0 can only be 4. – Tejay Lovelock May 23 '17 at 21:45

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