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I'm stuck on this problem:

Let a random vector $X$ be given in $\mathbb{R^{10}}$ with the standard scalar product. It is known that $\mathbb{E}[XX^T] = 5I_{10}$, $I_{10}$ – identity matrix of order 10. Let $t$ be a non-random vector of length 5 ($||t|| = 5$).

I need to find the value of this expression:

$$\mathbb{E}[X^TX] - \mathbb{E}\left(t^TX\right)^2$$

if I'm not mistaken then $\mathbb{E}[X^TX] = 5$. But I can't find the second term. Help me please

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  • $\begingroup$ What you wrote is not an equation. $\endgroup$
    – NN2
    Commented Sep 4, 2023 at 8:34
  • $\begingroup$ @NN2 thank you, you're right, this is an expression $\endgroup$
    – Andrew
    Commented Sep 4, 2023 at 8:53

1 Answer 1

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Denote $X = (X_1, ...,X_{10})$ and $t =(t_1,...,t_{10})$. From $\mathbb{E}[XX^T] = 5I_{10}$ and $||t|| = 5$ we have:

$$\mathbb{E}(X_iX_j)=5\cdot\mathbf{1}_{\{i=j \}}$$ $$\sum_{i=1}^{10} t_i^2 = 25$$

Then

$$\begin{align} L &:= \mathbb{E}[X^TX] - \mathbb{E}\left(t^TX\right)^2 \\ &=\mathbb{E}\left[\sum_{i=1}^{10}X_i^2\right] - \mathbb{E}\left[\left(\sum_{i=1}^{10}t_iX_i\right)^2\right]\\ &=\sum_{i=1}^{10}\mathbb{E}\left[X_i^2\right] - \sum_{i=1}^{10}\sum_{j=1}^{10}\mathbb{E}\left[t_it_jX_iX_j\right]\\ &=\sum_{i=1}^{10}\mathbb{E}\left[X_i^2\right] - \sum_{i=1}^{10}\mathbb{E}\left[t_i^2X_i^2\right]\\ &= 10\cdot 5 - \left(\sum_{i=1}^{10}t_i^2\right)\cdot5\\ &=-75 \end{align}$$

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