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A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a 6 and $Y$ the number of rolls needed to obtain a 5. I need help in computing $E(X |Y = 1)$ and $E(X|Y=5)$.

I know that $X$ is a geometric random variable, so $E(X)=6.$ My guess for the first one is 7, but I'm not sure. Given $Y=1$, it means that the first roll can't be a 6, so it will take at least 2 rolls to obtain a 6. Is my reasoning right?

jojo
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2 Answers2

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EDIT : Thanks to Andre, he gave me a slap in the face right when I needed it, i.e. before my applied analysis exam tomorrow. XD I'll make this answer right before I bring anymore shame on me.

My ex-approach for the case $Y=5$ is indeed making everything more complicated. The more easy and non-complicated way to do this is just use the definition of the expectation : $$ \mathbb E(X \, | \, Y = 5) = \sum_{n=1}^{\infty} \, n \, \mathbb P(X = n \, | \, Y = 5). $$ Now we just need to compute those probabilities. We know that $P(X = 5 \, | \, Y = 5) = 0$, so that removes this term. For the first four terms, note that $$ \mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^{n-1} \left( \frac 15 \right) = \frac{4^{n-1}}{5^n}, n=1, 2, 3, 4. $$ because the condition $Y=5$ only means for $X$ that the first four rolls cannot take the value $5$, hence the value of those rolls become independent and uniformly distributed over $\{1,2,3,4,6\}$.

For the next rolls, $Y=5$ gives information on the first five rolls but no information on the ones after. Thus $$ \mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^4 \left( \frac 56 \right)^{n-6} \left( \frac 16 \right). $$ The $n-6$ stands for the number of rolls after the first five rolls which are not a 6 before you actually get your first $6$ (which gives the $1/6$ term). Therefore, the series we look at in the expectation can be computed, after the first 5 terms, as the derivative of a geometric series in $\left( \frac 56 \right)$. I don't wanna compute it right now because I am going to mess it up, I am definitively too tired for this.

Patrick Da Silva
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  • Thanks Patrick: I got $\mathbb P(X < 5 , | , Y = 5) = \frac{369}{625}$. So $P(X > 5 , | , Y = 5)=\frac{265}{625}$. Are these right? I'm also finding difficulty in computing the expectations. Are those 4 and 10 respectively? – jojo Dec 19 '11 at 05:54
  • Sorry, I meant $P(X > 5 , | , Y = 5)=\frac{256}{625}$. If you can give a detailed solution too that'll be nice. Thanks. – jojo Dec 19 '11 at 06:07
  • I am a little busy right now, don't have time to explicitly compute. I trust Andre Nicolas, he's very usually right. =) – Patrick Da Silva Dec 19 '11 at 06:17
  • @AndréNicolas: Thanks. I figured my 10 should be 11, using the memoryless property. How do I get $\mathbb E(X , | , Y=5, X < 5)$? – jojo Dec 19 '11 at 06:29
  • @jojo : You can easily determine the distribution of $X , | , Y = 5, X < 5$ as a random variable. The condition $Y=5$ does not influence probabilities so only the condition $X<5$ is relevant. How do you determine the distribution of $X , | , X < 5$? Hint : $\mathbb P(X = i , | , X = 5) = \mathbb P(X=i \cap X < 5) / \mathbb P(X < 5)$. – Patrick Da Silva Dec 19 '11 at 06:39
  • Just calculate. Given the info $Y=5$, prob. that $X=1$ is $1/5$, prob. that $X=2$ is $(4/5)(1/5)$, and so on up to $X=4$. Use ordinary formula for expectation. That's one term. You could instead condition on $X<5$, but that just lengthens the calculation without structurally changing it. For other term, need to multiply $11$ by suitable factor. – André Nicolas Dec 19 '11 at 06:41
  • Both of you have been really helpful. Thanks. – jojo Dec 19 '11 at 06:50
  • @Andre : Not if we know that $X < 5$... I said OP needed to determine the distribution of $X , | , Y = 5, X < 5$. It clearly has the same distribution as $X , | , X < 5$. That's what I was saying. It sums up to "lengthening the computations" as you stated, but sometimes it is good to lengthen the computations to wander in safe lands when one is not quite used to tricky manoeuvers. =) That is why I used well-known formulas. – Patrick Da Silva Dec 19 '11 at 06:57
  • Ouch. I'm tired. Sorry. – Patrick Da Silva Dec 19 '11 at 07:11
  • I edited the entire answer... My idea was totally crappy, and a quick hot, waking-up shower made me realize how badly I was in the wrong direction. Sorry for that. And thanks to Andre. – Patrick Da Silva Dec 19 '11 at 08:05
  • @Patrick Da Silva: No big deal, just a minor misreading of the definition of random variable $Y$. – André Nicolas Dec 19 '11 at 14:54
  • @Andre : It was not a misreading, I read it correctly the first time, I was just so tired that along the way the definition of Y changed in my memory and I started saying dumb things. I really shouldn't answer when I'm tired... happened to me too often to give bad answers on this site when I was tired. – Patrick Da Silva Dec 19 '11 at 16:53
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E(X | Y = 1)= (1 * 1/5)+ ((6+1)*(4/5)) = 5.8

E(X | Y = 5)= (1 * 1/5)+ ((6+5)*(4/5)) = 9

Tino
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