Find the distribution of $Y_3=(Y_2|Y_1=3)$ and calculate $\mathbb E(Y_3)$

Question

Question: A random sample of size two is taken from the following distribution:
\begin{array}{|r|r|r|r|} \hline X & 0 & 1 & 2 & 3\\ \hline \mathbb P(X=x) &0.3 &0.4&0.2&0.1\\ \hline \end{array}
$(a)$ Find the distribution of $Y_1=X_1+X_2$
$(b)$ Find the distribution of $Y_2=X_1^2+X_2^2$
$(c)$ Find the distribution of $Y_3=(Y_2|Y_1=3)$ and calculate $\mathbb E(Y_3)$

My approach:
I am confused how to solve $(c)$. To make it easier I give what have I done .
$(a)$: the distribution of $Y_1=X_1+X_2$
\begin{array}{|r|r|r|r|r|r|r|r|} \hline Y_1 & 0 & 1 & 2 & 3& 4 & 5 & 6\\ \hline \mathbb P(Y_2=y) &0.09&0.24&0.28&0.22&0.12&0.04&0.01\\ \hline \end{array} $(b)$: the distribution of $Y_2=X_1^2+X_2^2$
\begin{array}{|r|r|r|r|r|r|r|r|} \hline Y_2 & 0 & 1 & 2 & 4 & 5 & 8 & 9 & 10 & 13 & 18 \\ \hline \mathbb P(Y_2=y) & 0.09 & 0.24 & 0.16 & 0.12 & 0.16 & 0.04 &0.06 &0.08&0.04&0.01 \\ \hline \end{array}
Any hint or solution will be appreciated . Thanks in advance .

Answer 1

Note that all involved random variables are discrete. Unfortunately, $Y_2$ and $Y_1$ are not independent, so we have to do some small calculations:

We have for all $y$ in the image of $Y_3$: $$\Bbb P(Y_3=y):=\frac{\Bbb P(Y_2=y \cap Y_1 = 3)}{\Bbb P(Y_1=3)}\approx4.55\cdot \Bbb P(Y_2=y \cap Y_1=3).$$

Note that $Y_1=3\iff X_1+X_2=3$ which is the case in exactly the following cases:

• $X_1=0,X_2=3$ (hence, $Y_2=9$),
• $X_1=1,X_2=2$ (hence, $Y_2=5$),
• $X_1=2,X_2=1$ (hence, $Y_2=5$),
• $X_1=3,X_2=0$ (hence, $Y_2=9$).

So $Y_3$ only takes the values $5, 9$. We have \begin{split}\Bbb P(Y_3=5)&\approx 4.55\cdot\big(\Bbb P(X_1=1\cap X_2=2)+\Bbb P(X_1=2\cap X_2=1)\big)\\&=9.1\cdot\Bbb P(X_1=1)\cdot\Bbb P (X_2=2)=9.1\cdot0.4\cdot0.2\\&=0.728.\end{split}

And, similarly, \begin{split} \Bbb P(Y_3=9)&\approx 9.1\cdot\Bbb P(X_1=0)\cdot\Bbb P (X_2=3) = 9.1\cdot0.3\cdot 0.1=0.273. \end{split}

From this, one can easily calculate $\Bbb E Y_3$.

(I am glad that my calculations got verified in the sense that $\Bbb P(Y_3=5)+\Bbb P(Y_3=9)\approx 1$.)