# Covariance of Sums of Random Variables/Lemma

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## Theorem

Let $n$ be a strictly positive integer.

Let $\sequence {X_i}_{1 \mathop \le i \mathop \le n}$ be a sequence of random variables.

Let $Y$ be a random variable.

Then:

- $\ds \cov {\sum_{i \mathop = 1}^n X_i, Y} = \sum_{i \mathop = 1}^n \cov {X_i, Y}$

## Proof

Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition:

- $\ds \cov {\sum_{i \mathop = 1}^n X_i, Y} = \sum_{i \mathop = 1}^n \cov {X_i, Y}$

### Basis for the Induction

We have that:

- $\ds \cov {\sum_{i \mathop = 1}^1 X_i, Y} = \cov {X_1, Y} = \sum_{i \mathop = 1}^1 \cov {X_i, Y}$

We therefore have that $\map P 1$ is true.

This is our basis for the induction.

### Induction Hypothesis

Suppose that $\map P n$ is true for some fixed $n \in \N$.

That is:

- $\ds \cov {\sum_{i \mathop = 1}^n X_i, Y} = \sum_{i \mathop = 1}^n \cov {X_i, Y}$

We aim to show that it logically follows that $\map P {n + 1}$ is true.

That is:

- $\ds \cov {\sum_{i \mathop = 1}^{n + 1} X_i, Y} = \sum_{i \mathop = 1}^{n + 1} \cov {X_i, Y}$

### Induction Step

This is our induction step:

We have:

\(\ds \cov {\sum_{i \mathop = 1}^{n + 1} X_i, Y}\) | \(=\) | \(\ds \cov {\sum_{i \mathop = 1}^n X_i + X_{n + 1}, Y}\) | splitting up the sum | |||||||||||

\(\ds \) | \(=\) | \(\ds \cov {\sum_{i \mathop = 1}^n X_i, Y} + \cov {X_{n + 1}, Y}\) | Covariance of Linear Combination of Random Variables with Another | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \cov {X_i, Y} + \cov {X_{n + 1}, Y}\) | Induction Hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^{n + 1} \cov {X_i, Y}\) |

Hence the result by induction.

$\blacksquare$