Distributions and expectation of DRVsAQA A-Level Further Maths: Revision notes

A discrete random variable (DRV) $X$ takes separate values $x_1,x_2,\ldots$ each with a probability. The probability distribution can be given as a table or as a function such as $P(X=x)=kx$ for $x=1,2,3,4$. Two rules always hold: $0\le P(X=x)\le1$ and $\sum P(X=x)=1$. The second is how you find an unknown constant: add all the probabilities, set the total equal to $1$ and solve. Example: $P(X=x)=kx$ for $x=1,2,3,4$ gives $k(1+2+3+4)=1$, so $k=\frac1{10}$ and the probabilities are $0.1,0.2,0.3,0.4$.

To find the probability of an event, add the probabilities of the values that satisfy it: $P(X\ge3)=P(X=3)+P(X=4)$. For a table, a cumulative column $P(X\le x)$ helps. The mode is the value with the greatest probability. The median is the smallest value $m$ with $P(X\le m)\ge0.5$. Example: for $P(X=1,2,3,4)=0.1,0.3,0.4,0.2$, the cumulative probabilities are $0.1,0.4,0.8,1$. The mode is $3$ and, because $P(X\le2)=0.4<0.5\le P(X\le3)$, the median is $3$.

The expected value (mean) of $X$ is the probability-weighted average: $E(X)=\sum x_ip_i.$ More generally $E(X^2)=\sum x_i^2p_i$: square the value, not the probability. Note that $E(X^2)\ne[E(X)]^2$ in general. Example: with probabilities $0.1,0.3,0.4,0.2$ on $1,2,3,4$: $E(X)=0.1+0.6+1.2+0.8=2.7$ and $E(X^2)=0.1+1.2+3.6+3.2=8.1$. A mean need not be a possible value of $X$: it is a long-run average.

The variance measures spread about the mean: $\mathrm{Var}(X)=E(X^2)-[E(X)]^2.$ It is the same as $\sum(x_i-\mu)^2p_i$, but the formula above is faster. The standard deviation is $\sqrt{\mathrm{Var}(X)}$ and has the same units as $X$. Example: $E(X)=2.7$, $E(X^2)=8.1$, so $\mathrm{Var}(X)=8.1-7.29=0.81$ and the standard deviation is $0.9$. Variance can never be negative; if you get a negative value, recheck $E(X^2)$.

Information about $E(X)$ gives a second equation. With two unknowns $p$ and $q$ you need two equations: one from $\sum P=1$ and one from $E(X)$. Example: $P(X=1,2,3,4)=0.2,p,q,0.3$ and $E(X)=2.6$. Sum: $p+q=0.5$. Mean: $0.2+2p+3q+1.2=2.6$, so $2p+3q=1.2$. Hence $q=0.2$ and $p=0.3$. For a function such as $P(X=x)=\frac kx$, write out each probability, find $k$ from the sum, then find $E(X)$, $E(X^2)$ and the variance as before. Here $E(X)=\sum x\cdot\frac kx=4k$ and $E(X^2)=10k$.