Conditional probability with max(X, Y) Let $Y_n=$ the outcome of the $n$-th die roll, let $X_{n+1} = \max \{X_n, Y_{n+1}\}$ with $X_1=Y_1$.
What is $P(X_{n+1}=j \ | X_1=i_1, ..., X_n=i)$?
I know that it is $P(\max \{X_n, Y_{n+1} \}=j \ | X_1=i_1, ..., X_n=i)$, and either $X_n$ is the maximum or $Y_{n+1}$ is.
Will then $$P(\max \{X_n, Y_{n+1} \}=j \ | X_1, ..., X_n) = P(X_n = j \ | X_1=i_1, ..., X_n=i) + P(Y_{n +1} = j \ | X_1=i_1, ..., X_n=i)?$$
If this was true, then the above would equal $$\frac{P( X_1=i_1, ...,X_n = j=i)}{P(X_1=i_1, ..., X_n=i)} + P(Y_{n +1} = j).$$
How can I simplify this?

I don't think the equation holds. In fact, we have
$$
P(\max\{X_n, Y_{n+1}\} = j\ |\ X_1 = i_1, \cdots, X_n = i_n) = P(\max\{X_n, Y_{n+1}\} = j | X_n = i_n)
$$
The presence of $X_n$ hides $X_1$, $X_2$, $\cdots, X_{n-1}$.

The value of $P(\max\{X_n, Y_{n+1}\} = j | X_n = i_n)$ depends on the values of $j$ and $i_n$:


*

*$j > i_n$


$P(\max\{X_n, Y_{n+1}\} = j | X_n = i_n)=P(Y_{n+1}=j | X_n = i_n) = \frac{1}{6}$, assuming the dice has 6 sides.


*

*$j = i_n$


$P(\max\{X_n, Y_{n+1}\} = j | X_n = i_n)=P(Y_{n+1}\leq j | X_n = i_n) = \frac{j}{6}$


*

*$j < i_n$


$P(\max\{X_n, Y_{n+1}\} = j | X_n = i_n)=0$
