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it is a problem from my exam praperation sheet

Let $U$ , $Y$ be independent random variables. Here $U$ is uniformly distributed on $(0 , 1)$ . whereas $Y$ is $\frac{1}{4} \delta(0) + \frac{3}{4} \delta (1)$ . Let $X := UY$ .

(a)Find $E[X]$

(b)Graph $F$ and find $P \left( 0 \le X \le \frac{2}{3} \right)$

can any one tell me how to solve this problem? Thanks

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You look for E(UY) where U and Y are independent. Does this ring any bell? – Did Jun 9 '12 at 16:52

For $E(X)$, one can note that since $Y$ and $U$ are independent, $E(X)=E(UY)=E(U)E(Y)$.

A fancier derivation which can be useful when we are dealing with something more complicated than sum or product is to use conditional expectation. We have $E(X|Y=0)=0$, and $E(X|Y=1)=E(U)=1/2$. Thus by what is sometimes called the Law of Total Expectation, $E(X)=(0)(1/4)+(1/2)(3/4)$.

For the second part, the event $X\le 2/3$ can happen in two ways: (i) $Y=0$ and $X\le 2/3$, or (ii) $Y=1$ and $X\le 2/3$.

(i) The conditional probability that $X\le 2/3$, given that $Y=0$, is $1$. (ii) The conditional probability that $X\le 2/3$, given that $Y=1$, is $2/3$. So our required probability is $$\frac{1}{4}\cdot 1 +\frac{3}{4}\cdot\frac{2}{3}.$$

Remark: It may be useful to tell a story about what is going on. We are thirsty, and end up playing the following game. An amount of water, uniformly distributed in $[0,1]$ (litres) is put into a jar. We then flip a coin that has probability $1/4$ of landing head, and $3/4$ of landing tail. If it lands head, we don't get the water in the jar. If it lands tail, we do get the water.

Then the amount of water we get has the same distribution as the random variable $UY$ of the question. So in the first part, we are calculating the mean amount of water we get, and in the second part we are calculating the probability that the amount of water we get is $\le \frac{2}{3}$ (of a litre).

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-1. E(X) cannot be both 0 and 1/2. (And 3/8...) – Did Jun 9 '12 at 16:58
@did: Thanks, was trying to sneak around conditioning. Can get away with sloppiness in simple cases such as this one, but it is a long run bad habit. – André Nicolas Jun 9 '12 at 18:29
Un-downvoted. On a different note, I find increasingly odd to see full-bolts-on solutions appear, where the OP would obviously rather need a reminder of some basics (here, that if independence then the expectation of a product is the product of the expectations). To see hints such as the one I gave in a comment made useless by some newbie eager to write any solution is only natural, I guess, but I find it less so when the solution is posted by an extremely experienced user. – Did Jun 9 '12 at 18:56
I have a leaning to detail. But it is certainly true that the product calculation could have been omitted, and will be. – André Nicolas Jun 9 '12 at 19:05

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