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Let us assume that we have a coin that is deformed and weighted. When you toss the coin, it lands on its side in 20% of the tosses. The coin lands on heads 30% of the time, and otherwise on tails. 1.We are now tossing our deformed, weighted coin several times until it lands on its side. Let us denote Y as the number of times the coin was actually tossed. Is Y a random variable? Explain your reasoning. 2. Assuming now that Y = 9 (where Y is from the previous assignment), what is the expected number of heads tossed before the coin landed on its side? Use the conditional expectation and explain your result.

1.I can say that y is a random variable because it is F-B(R) measurable but how to prove that it is F-B(R) measurable? 2. Let Y = no. Of toss X= no. Of heads tossed before the coin landed on its side. If now we compute E(X|Y=9) that means on 9th toss the coin landed on its side So we have X~B(n,p) with n=8 and p=0.3 E(X|Y=9) = np = 8x0.3 = 2.4 Is this approach right?

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  • $\begingroup$ You can only do that if you have already determined this $F$. Then by showing that for $n=1,2,\dots$ the sets $\{\omega\in\Omega\mid Y(\omega)=n\}$ are elements of $F$. If so then automatically for every Borel set $B$ the set $\{\omega\in\Omega\mid Y(\omega)\in B\}$ is an element of $F$ because it is a countable union of these sets. $\endgroup$ – drhab Feb 12 at 16:12
  • $\begingroup$ Please use MathJax $\endgroup$ – K.defaoite Feb 16 at 14:52
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Yes, you are almost there.

  1. $Y$ is a random variable and it is also a known rv: it's a geometric with density

$$f_Y(y)=0.8\cdot0.2^{y-1}$$

and CDF

$$F_Y(y)=1-0.2^y$$

$y=1,2,3,\dots$

Having defined a density (and a CDF too) it is enough as a proof.

  1. Given $Y=9$ the number of heads have a Binomial $Bin(8;0.3)$ as you stated
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