I have this problem which I have been struggling with for a while. I was hoping someone could help me figure out where my reasoning goes wrong.
Roll two dice and find the pmf of $X$ if $X$ is the smallest number.
To solve this, I introduce the random variables $X_1=\{1,2,...,6\}$ and $X_2=\{1,2,...,6\}$ which both denotes the variable for the two dices. Now i want to find $P(\min(X_1,X_2)=k)$ for $k=1,2,...,6$.
My plan is to try out different values for k and see if a pattern emerge. (Note that the events $X_1\geqq k$ and $X_2\geqq k$ are clearly independent.) $$P(\min(X_1,X_2)=1) = P(X_1\geqq 1 \cap X_2\geqq 1) = P(X_1\geqq 1)P(X_2\geqq 1) = \left(\frac{6}{6}\right)^2 = 1$$ $$P(\min(X_1,X_2)=2) = P(X_1\geqq 2 \cap X_2\geqq 2) = P(X_1\geqq 2)P(X_2\geqq 2) = \left(\frac{5}{6}\right)^2$$ and so on. I therefore think that the general expression for the pmf would be $$p_X(k) = P(\min(X_1,X_2)=k)=\left(\frac{7-k}{6}\right)^2 \text{ for } k=1,2,...,6$$
This is wrong, according to my text book the answer should be $$p_X(k) = \frac{1}{36} (13-2k) \text{ for } k=1,2,...,6.$$
I would really appreciate if someone could help me with this. I am completely new to probability theory. Thanks!
Your essential mistake is the assumption that: $$ \{\min(X_1,X_2)=k\}=\{X_1\geq k\}\cap\{X_2\geq k\}$$
The correct version is:
$$ \{\min(X_1,X_2)\geq k\}=\{X_1\geq k\}\cap\{X_2\geq k\}$$
so that for $k\in\{1,2,3,4,5,6,7\}$:
$$P(\min(X_1,X_2)\geq k)=P(X_1\geq k\wedge X_2\geq k)=P(X_1\geq k)P(X_2\geq k)=\left(\frac{7-k}6\right)^2$$
Then for $k\in\{1,2,3,4,5,6\}$: $$P(\min(X_1,X_2)=k)=P(\min(X_1,X_2)\geq k)-P(\min(X_1,X_2)\geq k+1)=$$$$\left(\frac{7-k}6\right)^2-\left(\frac{6-k}6\right)^2=\frac1{36}(13-2k)$$