Thread: independent variable distribution question

Hi, i am having trouble with a particular question and have no idea what to do. I am only just asking for a better understanding of the question but a simple solution for guidance would be very appreciative.

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At the start of their marriage, a couple who prefer girls to boys decide to keep having children until they have one more girl than boy. By various methods they can ensure that each time they conceive, the baby's gender is female with probability 2/3 and male with probability 1/3.

(a) Let X be the total number of children they end up having. By considering the gender of their first baby, show that
X = 1 with probability 2/3;
or
X = 1 + X1 + X2 with probability 1/3;
where X1;X2 are independent random variables each with the same distribution as X.

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Now for X1 and X2, if they have the same distribution would that mean they have the same expected value, variance and mgf etc. I understand that the first part is just the probability of conceiving a girl with a probability of 2/3 but the second part with the two independent variables confuses me. I am assuming that 1/3 is just the probability of getting a boy but don't understand how it works.

I am also suppose to find the variance and expected value but i think i might try to do them myself but any hints or guidance would be very helpful.

I am only trying to get a better understanding of the question such as the significance of the independent variables in X = 1 + X1 + X2.

Any advice or help would be very appreciative.

Can someone confirm that X = 1 + X1 + X2 with probability of 1/3 is just the probability of having more than one child since 2/3 is the probability of having one child. Could the variables X1 and X2 be the number of boys and girls but since they are suppose to have the same distribution, could it be possible or would the variables mean something else?