minor update to 2-d-1

_problems/unit-02/D-cdf-of-a-discrete-RV/1.md

@@ -53,7 +53,7 @@ To construct the pmf for $X$ let's compute the probability for possible value of
 
 As a check on this table, if represents a legitimate pmf for $X$, we see that for all values making up the supp($X$) the probabilities are greater than 0 and the sum of the probabilities is = 1.
 
-So, the soution to part 1 is
+So, the soution to **part 1** is
 
 $$
 \Rightarrow  X \in \\{ 0, 1, 2, 3, 4 \\} \text{, the pmf in bracket notation is: }
@@ -63,14 +63,14 @@ $$
 P(X=x) = \begin{cases} 0.0625, &x = 0\\\\ 0.25, &x = 1\\\\ 0.375, &x = 2\\\\ 0.25, &x = 3\\\\ 0.0625, &x = 4\\\\ 0, &otherwise\end{cases}
 $$
 
-For part 2 ...
+For **part 2** ...
 
 $$
 P(\text{ X is odd }) \Rightarrow P(\text{ X = 1 or X = 3 }) =
 \text{ (By Independence)  } P(X = 1) + P( X = 3)  = 0.5
 $$
 
-Finally for part 3 we can compute the cdf $\( F_{X}(x) \)$ for $X$ by recalling that
+Finally for **part 3** we can compute the cdf $\( F_{X}(x) \)$ for $X$ by recalling that
 
 $$ 
 \( F_{X}(x) \) :== P(X \leq x)
@@ -81,16 +81,16 @@ That is we sum all the mass from $\( -\infty \)$ to desired value of $x$ and rec
 $$
 F_{X}(x) = 
 \begin{cases}
-0 & \text{if } x < 0 \\\\
-0.0625 & \text{if } 0 \leq x < 1 \\\\
-0.3125 & \text{if } 1 \leq x < 2 \\\\
-0.6875 & \text{if } 2 \leq x < 3 \\\\
-0.9375 & \text{if } 3 \leq x < 4 \\\\
-1 & \text{if } 4 \leq x \\\\
+    0 & \text{if } x < 0 \\\\
+    0.0625 & \text{if } 0 \leq x < 1 \\\\
+    0.3125 & \text{if } 1 \leq x < 2 \\\\
+    0.6875 & \text{if } 2 \leq x < 3 \\\\
+    0.9375 & \text{if } 3 \leq x < 4 \\\\
+    1 & \text{if } 4 \leq x \\\\
 \end{cases} 
 $$
 
-For part 4 we can just read off of the cdf in bracket notation
+For **part 4** we can just read off of the cdf in bracket notation
 
 $$ 
 P(X \leq 2) \text{ is the sum of the probability mass values at 0, 1, 2 or from the bracket notation = } 0.6875